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muminat
3 years ago
9

It's emergency pls. ​

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

Answer:

see above there is the answer sorry for rough

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For the function defined by f(t)=2-t, 0≤t<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
. A recipe for cookies calls for 2/3 of a cup of sugar per batch of cookies. Elena used 5 and 1/3 cups of sugar to make multiple
Cerrena [4.2K]
Answer: 8 batches
Step-by-step explanation:
To get the number of batches she made, we can just use proportion to solve it. But first, we need to convert 5 1/3 to improper fraction
5 1/3 = 16/3
Then we can now use the proportion to solve
2/3 cups = 1 batch
16/3 cups = x (note:5 1/3=16/3)
cross multiply
2/3 × x = 16 /3
2x/ 3 = 16/3
we need to make x the subject of the formula, to do that we will multiply each side of the equation by 3/2
2/3 × 3/2 x = 16/3 × 3/2
6x /6 = 48 / 6
x = 8
Therefore she made 8 batches of cookies.
4 0
3 years ago
Read 2 more answers
Find the equation of the line passing through the points (5,21) and (-5,-29)
Assoli18 [71]

Answer:

Look below

Step-by-step explanation:

lets have -29 be y2 and -5 be x2

21 - (-)29

5 - (-)5

Subtracting a negative is just adding it so

21 + 29   50
5 + 5       10

50/10

this simplified, is 5/1 or 5

y=5x+b

now we need to find b, so we substitute  y and y

x = 5

y = 21

21=5*5=b

5*5 is 25

now we need to get to 21


25 -4 is 21

b = -4

y=5x-4

3 0
2 years ago
Please answer these questions
lianna [129]

Answer:

8.  -3, 16

9.  -3, 4.5

Step-by-step explanation:

See attached worksheet.

4 0
1 year ago
Read 2 more answers
Sam created a porch that was part wooden and part concrete, as shown in the figure below. Note: Figure not drawn to scale If a =
rjkz [21]
Youre gonna have to upload a picture of the porch, "figure below"
3 0
4 years ago
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