Can someone please help me solve this?
1 answer:
You might be interested in
The half-range sine series is the expansion for 
with the assumption that is considered to be an odd function over its full range, 
. So for (a), you're essentially finding the full range expansion of the function
with period 2 so that
for 
and integers 
.
Now, since is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with
where
In this case,
, so
The half-range sine series expansion for is then
which can be further simplified by considering the even/odd cases of, but there's no need for that here.
The half-range cosine series is computed similarly, this time assuming is even/symmetric across its full range. In other words, you are finding the full range series expansion for
Now the sine series expansion vanishes, leaving you with
where
for
. Again, . You should find that
Here, splitting into even/odd cases actually reduces this further. Notice that when is even, the expression above simplifies to
while for odd, you have
So the half-range cosine series expansion would be
Attached are plots of the first few terms of each series overlaid onto plots of. In the half-range sine series (right), I use 
terms, and in the half-range cosine series (left), I use 
or 
terms. (It's a bit more difficult to distinguish from the latter because the cosine series converges so much faster.)
with period 2 so that
Now, since
where
In this case,
The half-range sine series expansion for
which can be further simplified by considering the even/odd cases of
The half-range cosine series is computed similarly, this time assuming
Now the sine series expansion vanishes, leaving you with
where
for
Here, splitting into even/odd cases actually reduces this further. Notice that when
while for odd
So the half-range cosine series expansion would be
Attached are plots of the first few terms of each series overlaid onto plots of
Answer:
U = 16206 + 22.2x
Step-by-step explanation:
This year, water usage has increased by 20% for both regular days and special events.. Write an equation for the total water usage, U, at the park this year in terms of x, the number of special events the park has hosted
Solution:
Last year regular usage of water was 13505 gallons, where as it was 18.5 gallons per special event. Let us assume that there were x special invents, hence:
The total usage for last year = 13505 + 18.5x
This year, the total usage (U) increased by 20%, hence:
Total usage for this year (U) = total usage for last year + 20% of total usage for last year
U = (13505 + 18.5x) + 0.2(13505 + 18.5x)
U = 13505 + 18.5x + 2701 + 3.7x
U = 16206 + 22.2x
