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zysi [14]
3 years ago
14

Https://imgur.com/a/n3PRcMY

Mathematics
1 answer:
slava [35]3 years ago
8 0

Step-by-step explanation:

hakdogg happyly ever after

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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
3 years ago
Add. 4 7/12+2 3/4+7 5/12
mihalych1998 [28]

Answer:

C.14¾

3 9⁄10

Step-by-step explanation:

\frac{4}{5}  + 3 \frac{1}{10}  =  \frac{8}{10}  + 3 \frac{1}{10}  = 3 \frac{9}{10}

4 \frac{7}{12}  + 2 +  \frac{3}{4}  + 7 \frac{5}{12} =  13 \frac{12}{12}  +  \frac{3}{4}  = 14 +  \frac{3}{4}  = 14 \frac{3}{4}

I am joyous to assist you anytime.

3 0
3 years ago
A toy rocket is launched vertically from 5 feet above ground level with an initial velocity of 112 feet per second. The height h
Masja [62]

Answer:

Step-by-step explanation:

a)The height h after t seconds is given by the equation h(t)=-16t^2+112t+5.

Where 5 represents 5feet above the ground and this is the height from which the rocket was launched.

The equation is a quadratic equation. The plot of this equation on a graph would give a parabola whose vertex would be equal to the maximum height travelled by the rocket.

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = -16

b = 112

Vertex = - - 112/32= 3.5

So the rocket will attain maximum height at 3.5 seconds.

It will also take 3.5 seconds to reach the ground. This means it will take a total of 3.5 seconds + 3.5 seconds

= 7 seconds to hit the ground.

b) to find time it will take to be 100 feet above the ground,

-16t^2+112t-95 = 0

Using general quadratic equation formula,

a = -16

b = 112

c = -95

t = [-b+/-√b^2-4ac]/2a

t = [ -112 +/-√112^2-4×-16×-95]/16 × -2

= [-112 +/-√12544-6080]/-32

= (-112+80.4)/-32 or (-112-80.4)/-32

t = 0.9875 or t = 6.0125

So it will take 0.9875 or approximately 1 second to be 100 feet above the ground

7 0
3 years ago
Drag the slider to 2 inches. How many centimeters are in 2 inches? There are centimeters in 2 inches.
choli [55]

Answer: 5


Step-by-step explanation: There are 2.5 cm per in, so its just 2.5*2


4 0
2 years ago
Read 2 more answers
You want to put some money in a simple interest account it pays 8% interest annually for 2 years you would like to earn 500 doll
Semmy [17]

You deposit some money into a bank account paying 8% simple interest per year. You received $500 in interest after 2 years. How much the deposit (principal) was?


Result


The principal was $3125.


Explanation


Find principal by using the formula I=P⋅i⋅t, where I is interest, P is total principal, i is rate of interest per year, and t is total time in years.


In this example I = $500, i = 8% and t = 2 years, so


IPPP=P⋅i⋅t=Ii⋅t=5000.08⋅2=3125

4 0
3 years ago
Read 2 more answers
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