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Setler [38]
3 years ago
5

HELP ASAP PLEASE!!!!

Mathematics
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

q=(1,5) t=(-2,3)r=(3,-1)s=(0,0)

Step-by-step explanation:

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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
If (13,6) is an endpoint and (0,0)
TEA [102]

Answer:

its 123 because my mom said so and aslo my dog can fly like many people and stuff

5 0
3 years ago
Read 2 more answers
(-3y3-5y-2)+(-7y2+5y+2)
Slav-nsk [51]

Answer:

=-23y

Step-by-step explanation:

(-3y3-5y-2)+(-7y2+5y+2)

=-3yX3-5y-2-7yX2+5y+2

=-3X3y-5y-7X2y+5y-2+2

=-3X3y-7X2y-2+2

=-9y-7X2y-2+2

=-9y-14y-2+2

=-23y-2+2

=-23y

I hope this helps!

4 0
3 years ago
Read 2 more answers
Write 640 : 30 in the lowest form.
OLga [1]

Answer:

should be 64:3

Step-by-step explanation:

divide by 10

6 0
2 years ago
There are 10 red and 20 yellow balls. Drawing 3 balls without replacement, what is the chance the first is red and the next two
Xelga [282]

10 + 20 = 30 total balls.

Picking red = 10/30 = 1/3

After picking one ball there are 29 balls left.

Picking a yellow next is 20/29

Then there are 28 balls left and 19 yellow left, so picking another yellow would be 19/28

The probability of picking red first then two yellow = 1/3 x 20/29 x 19/28 = 95/609

6 0
3 years ago
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