# Book seems to be assuming too much

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• April 6th 2013, 12:36 AM
semouey161
Re: Book seems to be assuming too much
Quote:

Originally Posted by Paze
And just on a side-note. Is it because my function is designed in such a way that by limiting it to this extent, it just happens to converge on the maxima 1 and minima -1?

If so...If I am to identify another likewise function, do I need to derive it to get a correct minima/maxima result?

Let's be careful with our words here. What do you mean by converge? This word has a specific meaning in mathematics.

Next, if you are referring to your original function $\frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ , I think it has been made abundantly clear we are not doing anything with amplitude. With respect to the other function, the sin(stuff) function without the vertical shift and without the 25/4 in front of it, that one DOES have an amplitude of 1 (max y at 1 and min y at -1).

So far, I've been focusing on just the graph and behavior of the functions we talked about. I didn't even look to describe how to answer the questions.

My main thing right now is that you understand what's going on with the interval and the function in question.
• April 6th 2013, 10:49 AM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by semouey161
Here are the results I got:
Attachment 27826

Looks like the result on the left would be the minimum. Again, these results are only for the expression

$\sin\frac{2\pi(x-82)}{365}$ evaluated at x=1 and at x=365.

This means that your other expression $\frac{25}{4}\sin\frac{2\pi(x-82)}{365}+\frac{49}{4}$ has different values at x =1 and x =365 . Go ahead, plug in x = 1 and x = 365 into this one, and post the results. Also, I think I showed you the graph earlier for this expression on the interval in question. If I didn't, let me know and I'll get a picture of it here.

I understand that the interval of my function is 365. So for every 365 x, the function loops? That means that I must have a minima or a maxima at 365, right?

I should also make it clear that I fully understand the vertical shift and its effect on my function. I am only trying to understand the minima and maxima of my sin(stuff) part while the sin(stuff) part is limited to 1<x<365. You found the mimima by plugging in 1 in the equation. If I wanted to find a minima, I would differentiate it and find where it's derivative is 0, but I am guessing that if it is restricted, then you can plug in the lowest and highest values?
• April 6th 2013, 02:59 PM
emakarov
Re: Book seems to be assuming too much
Semouey, I am not sure you answered Paze's question, which is whether $h(x) = \sin(2\pi(x-82)/365)$ reaches 1 and -1 on the segment 1 ≤ x ≤ 365. As Paze correctly notes, sin(x/2) is defined on all real numbers and reaches 1 and -1, but when x is restricted to [0, 1] or even [0, 2π], sin(x/2) does not reach -1.

Quote:

Originally Posted by semouey161
To sum up, you have something*sin(stuff) + something, but the sin(stuff) still does its sine function thing, with a few alterations. Is the amplitude affected in just the sin(stuff) portion? No. Is its period affected? Yes. Does it still take on 1 and -1 if we limit the interval on which it is defined? No ... use a calculator and you'll see that we are at about -0.98 when x=1 and when x=365 .

In fact, h(x) does take on 1 and -1 between x = 1 and x = 365. Nobody said that h(1) = -0.98 is the minimum.

Quote:

Originally Posted by semouey161
Here are the results I got:
Attachment 27826

Looks like the result on the left would be the minimum.

No, h(1) is not the minimum. If the function is designed correctly, the minimum is reached around December 22, not on January 1.

Quote:

Originally Posted by Paze
So that I understand this correctly. If the period had been 366 then, with 365 days, I would not reach a whole period, and thus my minima and maxima would not be consistent with 1 and -1, correct?

If we restrict the domain so that its length (365) is less than the period (366), this may mean that not all values from the complete image (i.e., values taken by the function for all real arguments) are taken when the argument is in the restricted domain. As has beed said, sin(x/2) ≥ 0 for 0 ≤ x ≤ 2π, so negative values are not taken. However, one can see from the graph that h(x) still takes all values between -1 and 1 when 100 ≤ x ≤ 365.

Quote:

Originally Posted by Paze
The thing I am trying to wrap my head around is why we are so sure that it takes on 1 and -1 in my example when we limit x to 1-365.

This is answered by the following statement.

Theorem. Suppose $h:\mathbb{R}\to\mathbb{R}$ is a function with a persiod T, i.e., h(x + T) = h(x) for all x. Let I be the image of h(x), i.e., $I=\{h(x)\mid x\in\mathbb{R}\}$. For any number x0, denote D = [x0, x0 + T). Let h' be h restricted to D. Then h' : D → I is a surjection, i.e., h'(x) accepts every value in I for some x ∈ D.

Proof. Fix an arbitrary y ∈ I assumed at some x, i.e., h(x) = y. We need to find an x' ∈ D such that h'(x') = h(x') = y. Of course, if x ∈ D, then we can take x' = x and we are done. Suppose that x is located to the right of D, i.e., x ≥ x0 + T. Then we can repeatedly subtract T from x without changing the value of h and we will eventually reach a point x' inside D. We cannot "overshoot" D because the length of D, i.e., T, equals the step we are using to move left. More precisely, the number of steps we need to make is $n=\lfloor(x-x_0)/T\rfloor$, where $\lfloor\cdot\rfloor$ is the floor function, and x' = x - nT. Since h is periodic, h(x') = h(x). The fact that n ≤ (x - x0) / T < n + 1, which is the defining property of the floor function, implies that x0 ≤ x - nT < T + x0, i.e., x' ∈ D. The case when x lies to the left of D is considered similarly. (End of proof.)

The longest day is achieved for x such that 2π(x - 82) / 365 = π / 2, i.e., x = 365 / 4 + 82 = 173.25. Similarly, the shortest day is when 2π(x - 82) / 365 = 3π / 2, i.e., x = 3 * 365 / 4 + 82 = 355.75, which is around December 22.

A couple of remarks. We need to be careful with the boundary values of x. The length of the segment 1 ≤ x ≤ 365 is 364, which is less than the period, and so the value of the function at x = 1 and x = 365 is not exactly the same. The function should be considered either on [0, 365) or [1, 366). Second, the word "amplitude" is ambiguous. There is "peak-to-peak amplitude", which is 2 for sin(x), and simply "peak amplitude", which is 1 for sin(x).
• April 6th 2013, 06:34 PM
Paze
Re: Book seems to be assuming too much
Quote:

Originally Posted by emakarov
Semouey, I am not sure you answered Paze's question, which is whether $h(x) = \sin(2\pi(x-82)/365)$ reaches 1 and -1 on the segment 1 ≤ x ≤ 365. As Paze correctly notes, sin(x/2) is defined on all real numbers and reaches 1 and -1, but when x is restricted to [0, 1] or even [0, 2π], sin(x/2) does not reach -1.

In fact, h(x) does take on 1 and -1 between x = 1 and x = 365. Nobody said that h(1) = -0.98 is the minimum.

No, h(1) is not the minimum. If the function is designed correctly, the minimum is reached around December 22, not on January 1.

If we restrict the domain so that its length (365) is less than the period (366), this may mean that not all values from the complete image (i.e., values taken by the function for all real arguments) are taken when the argument is in the restricted domain. As has beed said, sin(x/2) ≥ 0 for 0 ≤ x ≤ 2π, so negative values are not taken. However, one can see from the graph that h(x) still takes all values between -1 and 1 when 100 ≤ x ≤ 365.

This is answered by the following statement.

Theorem. Suppose $h:\mathbb{R}\to\mathbb{R}$ is a function with a persiod T, i.e., h(x + T) = h(x) for all x. Let I be the image of h(x), i.e., $I=\{h(x)\mid x\in\mathbb{R}\}$. For any number x0, denote D = [x0, x0 + T). Let h' be h restricted to D. Then h' : D → I is a surjection, i.e., h'(x) accepts every value in I for some x ∈ D.

Proof. Fix an arbitrary y ∈ I assumed at some x, i.e., h(x) = y. We need to find an x' ∈ D such that h'(x') = h(x') = y. Of course, if x ∈ D, then we can take x' = x and we are done. Suppose that x is located to the right of D, i.e., x ≥ x0 + T. Then we can repeatedly subtract T from x without changing the value of h and we will eventually reach a point x' inside D. We cannot "overshoot" D because the length of D, i.e., T, equals the step we are using to move left. More precisely, the number of steps we need to make is $n=\lfloor(x-x_0)/T\rfloor$, where $\lfloor\cdot\rfloor$ is the floor function, and x' = x - nT. Since h is periodic, h(x') = h(x). The fact that n ≤ (x - x0) / T < n + 1, which is the defining property of the floor function, implies that x0 ≤ x - nT < T + x0, i.e., x' ∈ D. The case when x lies to the left of D is considered similarly. (End of proof.)

The longest day is achieved for x such that 2π(x - 82) / 365 = π / 2, i.e., x = 365 / 4 + 82 = 173.25. Similarly, the shortest day is when 2π(x - 82) / 365 = 3π / 2, i.e., x = 3 * 365 / 4 + 82 = 355.75, which is around December 22.

A couple of remarks. We need to be careful with the boundary values of x. The length of the segment 1 ≤ x ≤ 365 is 364, which is less than the period, and so the value of the function at x = 1 and x = 365 is not exactly the same. The function should be considered either on [0, 365) or [1, 366). Second, the word "amplitude" is ambiguous. There is "peak-to-peak amplitude", which is 2 for sin(x), and simply "peak amplitude", which is 1 for sin(x).

Thank you. Some of what you wrote was a bit out of my league but if I understand you correctly then we need to derive the function to prove its minima and maxima?
• April 7th 2013, 09:40 AM
emakarov
Re: Book seems to be assuming too much
Quote:

Originally Posted by Paze
if I understand you correctly then we need to derive the function to prove its minima and maxima?

I did not mention derivatives in my post at all. The function h' is not the derivative of h but the restriction of h to a single period. The point of the post is that a periodic function assumes all its values on a single period, which can start anywhere. In particular, the period of f(x) = sin(ax + b) is 2π / a, so f(x) assumes all values between -1 and 1 inclusive when x is restricted to any contiguous segment of length 2π / a.

We know where sin(x) reaches maxima and minima from the definition and not by differentiating it. Taking a derivative is just one way (though one of the most general) to find extrema of a function.

You can look up unfamiliar concepts from post #18, such as image of a function or set-builder notation $\{h(x)\mid x\in\mathbb{R}\}$, in Wikipedia, or feel free to ask about them here.
• April 9th 2013, 01:20 PM
Paze
Re: Book seems to be assuming too much
Ok. Thank you all for your help. I have learned a great deal from this discussion and I will continue to research your answers some more :)
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8670594692230225, "perplexity": 429.274880508217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398460942.79/warc/CC-MAIN-20151124205420-00030-ip-10-71-132-137.ec2.internal.warc.gz"}