How to solve the integral equation $$ \int_{-20}^{x} \left| \left| \left| \left| \left| \left| \left| \left| t \right| -1 \right| -1 \right| -1 \right| -1 \right| -1 \right| -1 \right| -1 \right| \,{\rm d}t={\frac {4027}{2}}?$$
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1graph this function to, find the answer – Khosrotash Nov 19 '14 at 20:08
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@ darya khosrotash: Could you kindly elaborate your comment? – Markiyan Hirnyk Nov 19 '14 at 20:11
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@MarkiyanHirnyk What have you tried ? What does the function's graph immediately show you ? – Hippalectryon Nov 19 '14 at 20:12
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@Hippalectryon: I think about that. The graph says nothing for me. – Markiyan Hirnyk Nov 19 '14 at 20:14
2 Answers
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$$\int_{-20}^{x}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt =\\\int_{-20}^{-7}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{-7}^{-5}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{-5}^{-3}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{-3}^{-1}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{--1}^{-1}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{+1}^{+3}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{+3}^{+5}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{+5}^{+7}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt+\\\int_{+7}^{x}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt=\frac{4027}{2}
$$so
$$ \int_{-20}^{-7}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt=(-20-(-7)\cdot(13)\cdot\frac{1}{2}=\frac{169}{2}\\\int_{-7}^{-5}||||||||t|−1|−1|−1|−1|−1|−1|−1|\,dt=2\cdot1\cdot\frac{1}{2}
$$so by simplify
$$\frac{169}{2} +7\cdot2\cdot1\cdot\frac{1}{2} + \frac{(x-7)(x-7)}{2}=\frac{4027}{2}\\ \frac{(x-7)(x-7)}{2}=\frac{4027-169-14}{2}
$$
Khosrotash
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If I don't make an error, the roots are $7+2\sqrt {962},7-2\sqrt{962}$ whereas Maple brings $69.$ – Markiyan Hirnyk Nov 19 '14 at 20:26
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Yes, you made an error; the second term of your simplified expression should have been $7\cdot2\cdot1\cdot\frac{1}{2}$ (not $5$). Count the humps in your camel-like diagram, there are 7 of them. The answer is $x=69$. – Mark Fischler Nov 19 '14 at 21:05
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The main tough part is getting a grasp on the function in the integral. Let $$ A_0(t) = |t| $$ and for $n > 0$ $$ A_n(t) = |A_{n-1}(t) - 1| $$ Then the function integrated is $A_7(t)$ (count the -1's).
Now it is easy to prove (by a two-step induction proof on odd and even $n$) that for all $n \in \Bbb{Z}$ $$ |t| \leq n \Longrightarrow A_{n}(t) = |t| - n $$ and $$ \int_{-n}^{n} A_{n}(t)dt = n $$
So, for $x > 7$, $$ \int_{-20}^{-7}A_7(t)dt = \int_{-20}^{-7} (-t-7) dt + 7 + \int_{7}^{x}(t-7)dt = \frac{169}{2} + 7 + \frac{(x-7)^2}{2} $$ Then $$\frac{183}{2} + \frac{(x-7)^2}{2}= \frac{4027}{2}$$ $$(x-7)^2 = 3844 $$ $$x = 62+7 = 69 $$ (The other solution, $x = -55$, clearly does not satisfy the original problem, since the integrand is always non-negative, so an integral from $=20$ to $-55 < -20$ is necessarily negative.)
Mark Fischler
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