1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
olganol [36]
2 years ago
8

Explain why it is helpful to know the basic function shapes and discuss some ways to remember them.

Mathematics
1 answer:
postnew [5]2 years ago
3 0
<h3>Explain why it is helpful to know the basic function shapes and discuss some ways to remember them. </h3>
  • Knowing the basic function shapes and discuss some ways to remember them is helpful because this is useful tools in the creation of mathematical models because we constantly make theories about the relationships between variables in nature and society. Functions in school mathematics are typically defined by an algebraic expression and have numerical inputs and outputs.
You might be interested in
For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
4 years ago
Three groups of students used different methods to estimate the diagonal length of a patio in feet. Their results were:
podryga [215]
F14.33 1425,413 that’s the right number
5 0
3 years ago
Ik this is easy but I just want to make sure if I’m right
Kipish [7]

Answer:

Fully color in 2 pentagons and then color only two triangles in the third pentagon

Step-by-step explanation:


5 0
4 years ago
Please helpppppppppppp
stira [4]
The answer would be 20
6 0
3 years ago
Write a word problem the equation -8f - 7 = -47
Ivenika [448]

Answer:

The weather at 5 pm is -7 degrees Celsius. Every hour, the temperature drops 8 degrees. How much time has gone by when the temperature is -47 degrees?

Step-by-step explanation:

The equation can be rearranged to get:

-7-8f=-47

Example: The weather at 5 pm is -7 degrees Celsius. Every hour, the temperature drops 8 degrees. How much time has gone by when the temperature is -47 degrees?

6 0
3 years ago
Read 2 more answers
Other questions:
  • What is the solution to the system of equations?
    7·1 answer
  • A loan with a balance of $210,000 prior to the June 1 payment was figured with interest at 5% annually and monthly principal and
    12·1 answer
  • VERY IMPORTANT! PLEASE ANSWER FULLY
    7·2 answers
  • In a Greek theater, the round flat space where the action took place was called the _____.
    15·2 answers
  • How many solutions are there to an inequality?
    10·1 answer
  • Helpppppppp<br><br><br><br><br> What is the solution to the equation below?
    12·1 answer
  • What is the solution <br><br> options:<br> a. (-2,-2)<br> b. (-1,-5)<br> c. (-5,-1)<br> d. (2,-1)
    5·1 answer
  • Given an angle that is 120°, find m3. *<br><br> O 60<br> O 100<br> O 120*
    9·2 answers
  • How to rewrite 3/4 and 7/9 so it has common denominators and then use &lt; or &gt; or = to order 3/4 and 7/9
    10·1 answer
  • H' (x, y) x-coordinate = ?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!