Math Challenge - September 2019

[fresh_42]

Spoiler: Speculating on question 15

These are speculations, I can't really answer the question.
$$
S = \int_{\frac{1}{2}}^{3} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx \, + \, \int_{\frac{1}{3}}^{2} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx\\
= \int_{\frac{1}{3}}^{3} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx+\int_{\frac{1}{2}}^{2} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx
$$
This seemed to be equal to zero so I tried computing the area using some triangles and rectangles. I think that it's because of the fact that $\log{\frac{1}{a}}+\log{a}=0$ or something like this.

So compute $I_n:=\displaystyle{\int_\frac{1}{n}^n} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx$.

 

[odd_even]

Spoiler: Number 15

Sorry, I am new here so I might get some of the formatting wrong.
I first expressed each integral into sums of two integrals:
$$\int_\frac 1 2^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx=\int_\frac 1 2^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx + \int_ 2^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$$$\int_\frac 1 3^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx=\int_\frac 1 3^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx - \int_ 2^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$Adding these two equations together:
$$S=\int_\frac 1 3^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx + \int_\frac 1 2^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$So, as said before, we need to compute the following integral:
$$I_n=\int_\frac 1 n^n \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$Let $u=\log {x}$:
$$\frac {du} {dx}=\frac 1 x$$$$e^u\,du=dx$$$$I_n=\int_{\log {\frac 1 n}}^{\log n} \frac {e^u} {\sqrt{e^{2u}+1}} {\frac u {\sqrt {e^u}}} \,du=\int_{-\log {n}}^{\log n} \frac {u{\sqrt{e^u}}} {\sqrt{e^{2u}+1}} \,du$$Multiply the numerator and denominator by $\frac {\sqrt{2}} {\sqrt{e^u}}$:$$I_n=\frac 1 {\sqrt{2}} \int_{-\log {n}}^{\log n} u {\sqrt{\frac 2 {e^u+e^{-u}}}} \,du=\frac 1 {\sqrt{2}} \int_{-\log {n}}^{\log n} \frac u {\sqrt{\cosh u}} \,du$$Observe that $\frac u {\sqrt{\cosh u}}$ is an odd function. Therefore: $$I_n=0$$ Finally: $$S=\int_\frac 1 3^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx + \int_\frac 1 2^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx=0+0=0$$

 

[fresh_42]

Spoiler: Number 15

Sorry, I am new here so I might get some of the formatting wrong.
I first expressed each integral into sums of two integrals:
$$\int_\frac 1 2^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx=\int_\frac 1 2^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx + \int_ 2^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$$$\int_\frac 1 3^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx=\int_\frac 1 3^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx - \int_ 2^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$Adding these two equations together:
$$S=\int_\frac 1 3^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx + \int_\frac 1 2^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$So, as said before, we need to compute the following integral:
$$I_n=\int_\frac 1 n^n \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx$$Let $u=\log {x}$:
$$\frac {du} {dx}=\frac 1 x$$$$e^u\,du=dx$$$$I_n=\int_{\log {\frac 1 n}}^{\log n} \frac {e^u} {\sqrt{e^{2u}+1}} {\frac u {\sqrt {e^u}}} \,du=\int_{-\log {n}}^{\log n} \frac {u{\sqrt{e^u}}} {\sqrt{e^{2u}+1}} \,du$$Multiply the numerator and denominator by $\frac {\sqrt{2}} {\sqrt{e^u}}$:$$I_n=\frac 1 {\sqrt{2}} \int_{-\log {n}}^{\log n} u {\sqrt{\frac 2 {e^u+e^{-u}}}} \,du=\frac 1 {\sqrt{2}} \int_{-\log {n}}^{\log n} \frac u {\sqrt{\cosh u}} \,du$$Observe that $\frac u {\sqrt{\cosh u}}$ is an odd function. Therefore: $$I_n=0$$ Finally: $$S=\int_\frac 1 3^3 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx + \int_\frac 1 2^2 \frac 1 {\sqrt{x^2+1}} {\frac {\log {x}} {\sqrt x}} \,dx=0+0=0$$

Well done! The integration itself can be done a bit easier, but you had all the right ideas. The problem should show, that beside the usual additive symmetry $\int_a^b + \int_b^a =0$ there is sometimes a multiplicative symmetry, too.

If you're interested in more of these, have a look at
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/

 

[benorin]

#8) $\sum_{n=0}^\infty \frac{(-1)^{n}}{\sqrt{n+1}}=\left( 1- \sqrt{2}\right) \zeta (\tfrac{1}{2} )$ hence convergent. But I bet you want the work, so: this a particular value of a Dirichlet series of the the form $\sum \tfrac{a_n}{n^z}$, and for such series there exists an abscissa of convergence $\lambda$ with the property that the series converges whenever $x>\lambda$ and diverges whenever $x<\lambda$-- Knopp, Theory and Application of Infinite Series, pg. 317, #7. Since it is well-known that the series $\zeta(s)= (1-2^{1-s})^{-1}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}$ converges for $\Re \left[ s\right] >0$, said abscissa is $\lambda =0$.

In the Cauchy product $\left(\sum_{n=0}^\infty \frac{(-1)^{n}}{\sqrt{n+1}}\right) ^2 =\sum_{j=0}^\infty \sum_{n=0}^j \tfrac{(-1)^{j}}{\sqrt{n+1}\cdot\sqrt{j-n+1}}$ consider the general term of the righthand side (the inner sum) $c_j :=\sum_{n=0}^j \tfrac{(-1)^{j}}{\sqrt{n+1}\cdot\sqrt{j-n+1}}$, by replacing each square root in the denominators with the largest such root, $\sqrt{n+1}$, we see that we must have

$$| c_j |\geq \frac{j+1}{\sqrt{j+1}\cdot\sqrt{j+1}}=1$$

and hence the given Cauchy product is certainly divergent. The reason is that the series we took the Cauchy product of is not absolutely convergent, hence not equal to the square of the sum.

 

[fresh_42]

#8) $\sum_{n=0}^\infty \frac{(-1)^{n}}{\sqrt{n+1}}=\left( 1- \sqrt{2}\right) \zeta (\tfrac{1}{2} )$ hence convergent. But I bet you want the work, so: this a particular value of a Dirichlet series of the the form $\sum \tfrac{a_n}{n^z}$, and for such series there exists an abscissa of convergence $\lambda$ with the property that the series converges whenever $x>\lambda$ and diverges whenever $x<\lambda$-- Knopp, Theory and Application of Infinite Series, pg. 317, #7. Since it is well-known that the series $\zeta(s)= (1-2^{1-s})^{-1}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}$ converges for $\Re \left[ s\right] >0$, said abscissa is $\lambda =0$.

In the Cauchy product $\left(\sum_{n=0}^\infty \frac{(-1)^{n}}{\sqrt{n+1}}\right) ^2 =\sum_{j=0}^\infty \sum_{n=0}^j \tfrac{(-1)^{j}}{\sqrt{n+1}\cdot\sqrt{j-n+1}}$ consider the general term of the righthand side (the inner sum) $c_j :=\sum_{n=0}^j \tfrac{(-1)^{j}}{\sqrt{n+1}\cdot\sqrt{j-n+1}}$, by replacing each square root in the denominators with the largest such root, $\sqrt{n+1}$, we see that we must have

$$| c_j |\geq \frac{j+1}{\sqrt{j+1}\cdot\sqrt{j+1}}=1$$

and hence the given Cauchy product is certainly divergent. The reason is that the series we took the Cauchy product of is not absolutely convergent, hence not equal to the square of the sum.

The Leibniz criterion as reason for convergence would have been a lot easier.