Puzzles
An integral
What is
$$\int_0^{\frac\pi2}\frac1{1+\tan^a(x)}\,dx?$$
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Surprisingly, the value of the integral does not depend on \(a\). To show this, first use \(\tan(x)=\frac{\sin(x)}{\cos(x)}\) on both integrals:
$$I_1=\int_0^{\frac\pi2}\frac1{1+\tan^a(x)}\,dx
$$$$=\int_0^{\frac\pi2}\frac{\cos^a(x)}{\cos^a(x)+\sin^a(x)}\,dx$$
$$I_2=\int_0^{\frac\pi2}\frac1{1+\cot^a(x)}\,dx
$$$$=\int_0^{\frac\pi2}\frac{\sin^a(x)}{\cos^a(x)+\sin^a(x)}\,dx$$
Now consider \(I_1+I_2\):
$$I_1+I_2 = \int_0^{\frac\pi2}\frac{\cos^a(x)+\sin^a(x)}{\cos^a(x)+\sin^a(x)}\,dx$$$$=\int_0^{\frac\pi2}1\,dx$$$$=\frac\pi2$$
By substituting \(u=\tfrac\pi4-x\) into \(I_2\), it can be shown that \(I_1 = I_2\).
Therefore \(I_1=\tfrac\pi4\).
Extension
What is
$$\int_0^{\frac\pi2}\frac1{1-\tan^a(x)}\,dx?$$
Find them all
Find all continuous positive functions, \(f\) on \([0,1]\) such that:
$$\int_0^1 f(x) dx=1\\
\mathrm{and }\int_0^1 xf(x) dx=\alpha\\
\mathrm{and }\int_0^1 x^2f(x) dx=\alpha^2$$
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$$0=\alpha^2-2\alpha^2+\alpha^2\\
=\int_0^1 x^2f(x) dx-2\int_0^1 \alpha xf(x) dx+\int_0^1 \alpha^2f(x)dx\\
=\int_0^1 (x^2-2\alpha x+\alpha^2)f(x)dx\\
=\int_0^1 (x-\alpha)^2 f(x)dx$$
\(f(x)\) and \((x-\alpha)^2\) are both positive so this is only possible if one of them if always zero.
But \((x-\alpha)^2\) is only zero when \(x=\alpha\) and \(\int_0^1 f(x) dx=1\) so \(f(x)\) cannot always be zero. Therefore no such function exists.
Extension
Find all continuous positive functions, \(f\) on \([0,1]\) such that:
$$\int_0^1 f(x) dx=1\\
\int_0^1 xf(x) dx=1\\$$
Integrals
$$\int_0^1 1 dx = 1$$
Find \(a_1\) such that:
$$\int_0^{a_1} x dx = 1$$
Find \(a_2\) such that:
$$\int_0^{a_2} x^2 dx = 1$$
Find \(a_n\) such that (for \(n>0\)):
$$\int_0^{a_n} x^n dx = 1$$
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$$1=\int_0^{a_1} x dx=\frac{a_1^2}{2}$$
So, \(a_1=\sqrt{2}\).
$$1=\int_0^{a_2} x^2 dx=\frac{a_2^3}{3}$$
So, \(a_2=3^{\frac{1}{3}}\).
$$1=\int_0^{a_n} x^n dx=\frac{a_n^{n+1}}{n+1}$$
So, \(a_n={(n+1)}^{\frac{1}{n+1}}\).
Extension
Find \(b_n\) such that (for \(n>1\)):
$$\int_{b_n}^{\infty} x^{-n} dx = 1$$
Double derivative
What is
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$
when:
(i) \(y=x\)
(ii) \(y=x^2\)
(iii) \(y=x^3\)
(iv) \(y=x^n\)
(v) \(y=e^x\)
(vi) \(y=\sin(x)\)?
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(ii) Differentiating \(y=x^2\) with respect to \(x\) \(\frac{dy}{dx}=2x\). Let \(g=\frac{dy}{dx}\). By the chain rule:
$$\frac{dg}{dy}=\frac{dg}{dx}\frac{dx}{dy}$$
$$=2\frac{1}{2x}$$
$$=\frac{1}{x}$$
So \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{1}{x}\)
(iii) By the same method, \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{2}{x}\)
(iv) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{n-1}{x}\)
(v) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=1\)
(vi) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=-\tan(x)\)
Extension
What is
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)$$
when \(y=f(x)\)?
Differentiate this
$$f(x)=e^{x^{ \frac{\ln{\left(\ln{x}\right)}}{ \ln{x}}} }$$
Find \(f'(x)\).
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$$f(x)=e^{x^{ \frac{\ln{\left(\ln{x}\right)}}{ \ln{x}}} }$$
$$=e^{e^{ \frac{\ln{\left(\ln{x}\right)}}{ \ln{x}}\ln{x}} }$$
$$=e^{e^{ \ln{\left(\ln{x}\right)}} }$$
$$=e^{\ln{x} }$$
$$=x$$
Therefore:
$$f'(x)=1$$
x to the power of x again
Let \(y=x^{x^{x^{x^{...}}}}\) [\(x\) to the power of (\(x\) to the power of (\(x\) to the power of (\(x\) to the power of ...))) with an infinite number of \(x\)s]. What is \(\frac{dy}{dx}\)?
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\(y=x^{x^{x^{x^{...}}}}\) so \(y=x^y=e^{y\ln{x}}\).
By the chain and product rules, \(\frac{dy}{dx}=e^{y\ln{x}}(\frac{dy}{dx}\ln{x}+\frac{y}{x})\).
Rearranging, we get \(\frac{dy}{dx}=\frac{ye^{y\ln{x}}}{x(1-e^{y\ln{x}}\ln{x})}\).
This simplifies to \(\frac{dy}{dx}=\frac{x^{x^{x^{x^{...}}}}x^{x^{x^{x^{...}}}}}{x(1-x^{x^{x^{x^{...}}}}\ln{x})}\).
Extension
What would the graph of \(y=x^{x^{x^{x^{...}}}}\) look like?
