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Miscellaneous exercises
(extra-curricular topic)
Approximation methods play quite an important role in solving equations and inequalities, especially in practice. Let us solve a few exercises like these. Let us first get familiar with an approximation process that can be used well in a seemingly simple example: halving the interval.The properties of the functions can often be use well while solving trigonometric equations or inequalities. The following few exercises give examples for this.
Example 1
Let us calculate the length of the sides of a square with an area of \latex{ 3 } units.
\latex{ t }
\latex{ x }
\latex{ t=3 }
Solution
If x denotes the length of the side of the square, then we want to find the positive solution of the equation \latex{x^2 = 3}. It is obviously the irrational number \latex{x = \sqrt3}.
This form cannot be used in practice; let us calculate the value of \latex{\sqrt3} to two decimal places. For example we can act as follows. Let us rewrite the equation in the following form:
This form cannot be used in practice; let us calculate the value of \latex{\sqrt3} to two decimal places. For example we can act as follows. Let us rewrite the equation in the following form:
\latex{x^2-3=0},
and let us plot the graph of the function \latex{f:x\mapsto x^2-3,\;\;x\geq0.} (Figure 25)
We want to find for which x will the value of f be 0. The function f is strictly increasing on its domain. \latex{f(1) = –2,\;\; f(2) = 1}, so the zero in question is in the interval \latex{\left[1; 2\right]}.
We want to find for which x will the value of f be 0. The function f is strictly increasing on its domain. \latex{f(1) = –2,\;\; f(2) = 1}, so the zero in question is in the interval \latex{\left[1; 2\right]}.
It is plausible to check the value of \latex{ f } at the midpoint of the interval: \latex{f\left(\frac{3}{2}\right)=-\frac{3}{4}} it is negative here, so the zero is in the interval \latex{\left[\frac{3}{2}; 2\right]}. At the midpoint of this \latex{f\left(\frac{7}{4}\right)=\frac1{16}} so it is positive, thus the zero is in the interval \latex{\left[\frac{3}{2}; \frac{7}{4}\right]}.
We can continue with this halving the interval process indefinitely. After the next \latex{ 3 } steps the following is obtained: \latex{f\left(\frac{111}{64}\right)=\frac{33}{4,096}=0.008} so the zero in question to \latex{ 2 } decimal places is:
We can continue with this halving the interval process indefinitely. After the next \latex{ 3 } steps the following is obtained: \latex{f\left(\frac{111}{64}\right)=\frac{33}{4,096}=0.008} so the zero in question to \latex{ 2 } decimal places is:
\latex{\frac{111}{64}=1.73}.
\latex{y=x^2-3}
\latex{ 1 }
\latex{ 0 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ 1 }
\latex{ 2 }
Figure 25
The idea behind the “halving the interval” method applied is usually formulated in a funny way as follows: How does a mathematician catch a lion in the Sahara desert?
Example 2
Let us solve the equation \latex{\log x = 0.1x} by using the corresponding graphs of functions. Let us give the roots with a precision of \latex{10^{–2}}.
Solution
Let us plot the graphs of the following functions in one coordinate system (Figure 26):
\latex{f:\R^+\rightarrow\R,\;\;f(x)=\log x} and \latex{g:\R^+\rightarrow\R,\;\;g(x)=0.1x}.
\latex{y=\log x}
\latex{y=0.1x}
\latex{ 1 }
\latex{ 2 }
\latex{ 10 }
\latex{ x }
\latex{ 1 }
\latex{ -1 }
\latex{ y }
Figure 26
The logarithm function with base ten is strictly increasing and concave on \latex{]0; +\infty[}. The image of the function \latex{ g } is a ray; it intersects the graph of \latex{ f } at two places at most. In the figure it can be seen that one of the intersection points belongs to the place \latex{x = 10}, here \latex{f(10) = g(10) = 1}. So one of the roots of the equation is \latex{x = 10}, this value is exact.
The other intersection point is between \latex{ 1 } and \latex{ 2 }, as it can be seen in the figure, because \latex{f(1) = 0 \lt g(1) = 0.1} and \latex{f(2) = 0.30 \gt g(2) = 0.2}.
The more precise approximation, for example with halving the interval, shows that the other root is \latex{ 1.37 } to \latex{ 2 } decimal places, as \latex{f(1.37) = 0.1367} and \latex{g(1.37) = 0.137}.
The more precise approximation, for example with halving the interval, shows that the other root is \latex{ 1.37 } to \latex{ 2 } decimal places, as \latex{f(1.37) = 0.1367} and \latex{g(1.37) = 0.137}.
Example 3
Let us solve the following equation to two decimal places: \latex{\cos x=x}.
Solution
Let us plot the graphs of the following two functions (Figure 27):
\latex{f:\R\rightarrow\R,\;\;f(x)=\cos x} and \latex{g:\R\rightarrow\R,\;\;g(x)=x}.
From the figure it can be seen that the equation has a root between \latex{ 0 } and \latex{\frac{\pi}{2}}.
Indeed, on the interval \latex{\left[0; \frac{\pi}{2}\right]} \latex{ g } is strictly increasing, \latex{ f } is strictly decreasing, and \latex{f(0)=1\gt g(0)=0,\;\;f\left(\frac{\pi}{2}\right)=0\lt g\left(\frac{\pi}{2}\right)=\frac{\pi}{2},} thus it must have exactly one root here. And if \latex{x \gt\frac{\pi}{2}}, then \latex{f(x) \gt 1} and \latex{f(x) \gt 1}, therefore after \latex{\frac{\pi}{2}} there are no roots. On the interval \latex{[-\frac{\pi}{2}; 0]} \latex{ f } is positive, \latex{ g } is negative, so there is no root here either, and if \latex{x\lt-\frac{\pi}{2},} then \latex{g(x)\lt-1,} \latex{f(x)\geq -1,} there is no root here either.
Indeed, on the interval \latex{\left[0; \frac{\pi}{2}\right]} \latex{ g } is strictly increasing, \latex{ f } is strictly decreasing, and \latex{f(0)=1\gt g(0)=0,\;\;f\left(\frac{\pi}{2}\right)=0\lt g\left(\frac{\pi}{2}\right)=\frac{\pi}{2},} thus it must have exactly one root here. And if \latex{x \gt\frac{\pi}{2}}, then \latex{f(x) \gt 1} and \latex{f(x) \gt 1}, therefore after \latex{\frac{\pi}{2}} there are no roots. On the interval \latex{[-\frac{\pi}{2}; 0]} \latex{ f } is positive, \latex{ g } is negative, so there is no root here either, and if \latex{x\lt-\frac{\pi}{2},} then \latex{g(x)\lt-1,} \latex{f(x)\geq -1,} there is no root here either.
On the interval \latex{[0; \frac{\pi}{2}]} with a more precise calculation we can find that the root in question to two decimal places is:
\latex{ 0.74 }.
\latex{y=\cos x}
\latex{y= x}
\latex{\frac{\pi}{2}}
\latex{-\frac{\pi}{2}}
\latex{ x }
\latex{ 1 }
\latex{ 1 }
\latex{ -1 }
\latex{ -1 }
\latex{ f }
\latex{ y }
Figure 14
It can be checked either with a pocket calculator or with a function value table.
We can conduct an interesting experiment with a calculator. Let us calculate the value of \latex{ cos1 } and then the value of the cosine function for it again, i.e. the value of \latex{cos (cos1)} , and so on. For this it is enough to keep pressing the “cos” button after the second step. After a few steps we will see that the calculator keeps “jumping” around the value
We can conduct an interesting experiment with a calculator. Let us calculate the value of \latex{ cos1 } and then the value of the cosine function for it again, i.e. the value of \latex{cos (cos1)} , and so on. For this it is enough to keep pressing the “cos” button after the second step. After a few steps we will see that the calculator keeps “jumping” around the value
\latex{ 0.739 },
after approx. \latex{ 30–35 } steps it will steadily show the value
\latex{ 0.7390851 }.
This method, called iteration, can also be applied many times when solving equations in the form \latex{f(x) = x}.
◆ ◆ ◆
UNCLE GEORGE!
NO ZITHERATION!
NO ZITHERATION!
Now we are going to solve some inequalities for which the properties of the functions appearing (for example increase, decrease, domain, range, etc.) play an important role during the solution. The graphs of the functions also give much assistance when solving.
Example 4
Let us solve the following inequality on the set of real numbers:
\latex{\log_{\frac{1}{2}}\left(\log_2\frac{x}{x+1}\right)\gt0.}
Solution
The function \latex{\R^+\rightarrow\R,\;\;x\mapsto\log_{\frac{1}{2}}x} is strictly decreasing; its graph can be seen in Figure 28.
The function takes positive values between \latex{ 0 } and \latex{ 1 }, so the inequality is equivalent to the double inequality
The function takes positive values between \latex{ 0 } and \latex{ 1 }, so the inequality is equivalent to the double inequality
\latex{0\lt \log_{2}\frac{x}{x+1}\lt 1.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \tag 1}
The function \latex{\R^+\rightarrow\R,\;\;x\mapsto\log_2x} is strictly increasing (Figure 29), it takes the values between \latex{ 0 } and \latex{ 1 } on the interval \latex{]1; 2[}.
\latex{y=\log_{\frac 1 2}x}
\latex{ 1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
Figure 28
\latex{y=\log_{2}x}
\latex{ 1 }
\latex{ 1 }
\latex{ 2 }
\latex{ y }
\latex{ x }
Figure 29
So the inequality (1) is equivalent to the inequality
\latex{1\lt\frac{x}{x+1}\lt2.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \tag 2}
As \latex{\frac{x}{x+1}=1-\frac{1}{x+1},} it is also worth examining the function
\latex{\left(\R\backslash\left\{-1\right\}\right)\rightarrow\R,\;\;x\mapsto1-\frac{1}{x+1};}
its graph can be seen in Figure 30.
The function is strictly increasing on the interval \latex{]–\infty; –1[}; its values fill up the interval \latex{]1; +\infty[}. It is also increasing on the interval \latex{]–1; +\infty[}; its values give the interval \latex{]–\infty; 1[}. The function takes the value of 2 at the place –2. So the solutions of the inequality (2) are the real numbers \latex{x \lt –2}. These also give the solutions of the original inequality.
\latex{y=1-\frac{1}{x+1}}
\latex{ 2 }
\latex{ 1 }
\latex{ -1 }
\latex{ -2 }
y
x
Figure 30
Example 5
Let us solve the inequality \latex{\log_2\left(\sin{\frac{x}{2}}\right)\lt-1} on the set of real numbers.
Solution
Let us again use the functions which can be defined with the expressions appearing in the inequality.
The function \latex{\R^+\rightarrow\R,\;\;x\mapsto\log_2x}
\latex{ x } is strictly increasing (Figure 31), it takes the value of \latex{ –1 } at the place \latex{\frac{1}{2}}, so the inequality given is equivalent to the inequality
The function \latex{\R^+\rightarrow\R,\;\;x\mapsto\log_2x}
\latex{ x } is strictly increasing (Figure 31), it takes the value of \latex{ –1 } at the place \latex{\frac{1}{2}}, so the inequality given is equivalent to the inequality
\latex{0\lt\sin{\frac{x}{2}}\lt\frac{1}{2}.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\tag 1}
The function \latex{\R\rightarrow\R,\;\;x\mapsto\sin{\frac{x}{2}}} is periodic with the period \latex{4\pi} (Figure 32);
within one period it can be seen from the graph of the function for which values of \latex{ x } holds the inequality (1). So all the solutions are as follows:
within one period it can be seen from the graph of the function for which values of \latex{ x } holds the inequality (1). So all the solutions are as follows:
\latex{4k\pi\lt x\lt\frac{\pi}{3}+4k\pi\;\;(k\in\Z)} and \latex{\frac{5\pi}3+4n\pi\lt x\lt2\pi+4n\pi\;\;(n\in\Z).}
\latex{y=\log_2x}
\latex{\frac 1 2}
\latex{ 1 }
\latex{ x }
\latex{ -1 }
\latex{ y }
Figure 31
\latex{y=\sin{\frac x 2}}
\latex{-\pi}
\latex{-2\pi}
\latex{\frac{\pi}3}
\latex{\pi}
\latex{\frac{5\pi}3}
\latex{2\pi}
\latex{ 1 }
\latex{ -1 }
\latex{ y }
\latex{ x }
Figure 32
Exercises
{{exercise_number}}. Solve the following inequalities on the set of real numbers:
- \latex{\log_{\frac1 3}\left(\log_4\left(x^2-5\right)\right)\gt0};
- \latex{\log_2\frac{3x-1}{2x-1}\lt1};
- \latex{\log_2(\cos{2x})\lt-\frac{1}{2}}.
{{exercise_number}}. Solve the following equations with the help of graphical tools:
- \latex{10^{-x}=x};
- \latex{x=1+\frac1 2\times\sin{x}};
- \latex{\cot x=x,\;\;0\lt x\lt\pi}.
- \latex{2x-2=\sin x}.
