1/06/2023

Horizontal Stretches and Compressions of Functions

What happen if we consider changes to the inside of a function? As the figure blow shows that when we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function.


WHY ?  Let's take a closer look at the graph. Image that if inside the f(x), the x, is running 2 times faster or 2 times larger than the original x, the output f(x) will be 2x larger ?? The answer is NO. However, it will go 2 times faster to arrive the output that originally f(x) will reach. For example, for each output f(x), f(2x) only need 1/2 x to reach the output that f(x) does. In short, For each of the same output f(x), the inside change x, 2x, or 0.5x decide how fast the output f(x) will be reached.





Reference
Abramson, J. (2017). Algebra and trigonometry. OpenStax, TX: Rice University. Retrieved from https://openstax.org/details/books/algebra-and-trigonometry



Vertical Stretches and Compression of Functions

When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. But why?


That's because each of the same input have been changed to the output 2x or o.5 of the original functions.




Reference

Abramson, J. (2017). Algebra and trigonometry. OpenStax, TX: Rice University. Retrieved from https://openstax.org/details/books/algebra-and-trigonometry

1/05/2023

Arc Length, Radius, Radians, and Trigonometric

 1.  Find the length of an arc in a circle of radius 10 centimeters subtended by the central angle of 50°

An arc length S is the length of the curve along the arc. Just as the full circumference of a circle always has a constant ratio to the radius, the arc length produced by any given angle also has a constant relation to the radius, regardless of the length of the radius. The radian measure is also depends only on the angle. A full revolution (360°) equals 2π radians. A half revolution (180°) is equivalent to π radians.

Convert the 50° to radians 
  \frac{50^\circ }{360^\circ} = \frac{ \theta}{2 \pi }
  \theta= \frac{5 \pi }{18}
 S=r \theta=10( \frac{5 \pi }{18}) = \frac{25 \pi }{9} \approx8.72664



2.  Graph f(x)=x\ sin\ x on [-4π, 4π] and verbalize how the graph varies from the graphs of f(x)= \pm x




A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. One of the general forms of sinusoidal functions is y = Asin(Bx − C) + D, where A represents its Amplitude, B represents its Periods, C and D represents its shifts. In this case,  f(x)=xsinx  has only inconstant Amplitude X. Therefore, this function is expected to be a function that has the same period of  y=sinx   but with larger and larger Amplitudes. And, since the function  y=sinx  is repeating its values between the interval [-1, 1], when  sinx=1   or  sinx=-1  , the function  y=xsinx  will repeatedly has the value of X and -X. In fact, they are the intersections of these functions,  y=xsinx  ,  y=x  , and  y=-x

Graph f(x)= \frac{sin\ x}{x}  on the window [−5π, 5π] and describe freely what the graph shows.



Similar to the previous sinusoidal functions. One of the general forms of 
sinusoidal functions is y = Asin(Bx − C) + D, where A represents its Amplitude, B represents its Periods, C and D represents its shifts. This time,   \frac{sinx}{x}  has only inconstant Amplitude   \frac{1}{x}  Therefore, this function is expected to be a function that has the same period of  y=sinx   but with smaller and smaller Amplitudes. And, since the function  y=sinx  is repeating its values between the interval [-1, 1], when  sinx=1  or  sinx=-1  , the function  \frac{sinx}{x}  will repeatedly has the value of 1/X and -1/X. In fact, they are the intersections of these functions,  \frac{sinx}{x}  ,  \frac{1}{x}  , and
 \frac{-1}{x}




3. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 

If the ladder is lean against the TOP of the building and the building is vertically standing above the ground, we can use the sine function to solve this question. That is the heigh of the building is  23sin80^\circ \approx22.6505

If the ladder is NOT lean against the TOP of the building and the building is vertically standing above the ground, then the height we just calculated is just a portion of the building and it represents where the ladder reach up the side of the building.


1/04/2023

How can De Moivre's theorem be described? What is the scope of this theorem? Examples for roots and powers.

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. 

According to the De Moivre’s Theorem 

If Z = r(cos θ + isin θ) is a complex number, then Z= rn[cos(nθ) + isin(nθ)] 

where n is a positive integer. Zn = rn cis(nθ) 

How does it come from? To understand this theorem, we must know the products of complex numbers in polar form and the quotients of complex numbers in polar form first.

Recall that : 

    sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

    sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

    cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

    cos(α – β) = cos(α)cos(β) + sin(α)sin(β)


 


 (1+ \sqrt[]{3}i)^3

 r= \sqrt[]{ 1^{2}+( \sqrt[]{3})^2 } =2

 tan \theta= \sqrt[]{3}   ,   3(\frac{ \pi }{3})= \pi

 (1+ \sqrt[]{3}i)^3=2^3(cos\pi +sin \pi)=-8


 (\sqrt[]{2}+ \sqrt[]{2}i)^3

 r= \sqrt[]{ (\sqrt[]{2})^{2}+( \sqrt[]{2})^2 } =2

 tan \theta= 1  ,   \theta= \pi/4

 (\sqrt[]{2}+ \sqrt[]{2}i)^3=2^3(cos3\pi/4 +sin3\pi/4)=\sqrt[]{2}(-4+4i)


 (1+ \sqrt[]{3}i)^{ \frac{1}{2}}

 r= \sqrt[]{(1)^2+ (\sqrt[]{3})^2 } =2

 tan \theta= \sqrt[]{3}  ,    \theta= \frac{ \pi }{3}

 (1+ \sqrt[]{3}i)^{ \frac{1}{2}}   =(2^{ \frac{1}{2} })(cos(\frac{1}{2})(\frac{ \pi }{3})+isin(\frac{1}{2})(\frac{ \pi }{3}))  =   \sqrt[]{2} ( \frac{\sqrt[]{3}}{2} + \frac{1}{2} i)


  (\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}

 r= \sqrt[]{( \sqrt[]{2})^2+(\sqrt[]{2})^2} =2

 tan \theta=1   ,   \theta= \frac{ \pi }{4}

 (\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}   = \sqrt[]{2}(cos \frac{ \pi }{8} +isin \frac{ \pi }{8})= \sqrt[]{2}( \sqrt[]{\frac{1+ \frac{\sqrt[]{2}}{2} }{2} } +\sqrt[]{\frac{2-\sqrt[]{2} }{2} })

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