HELP ME ASAP PLEASEEE!!!!
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5.
Clearly, each even-order derivative will vanish, and the terms that remain will alternate in sign, so the Taylor series is given by
Your answer is off by a sign - the source of this error is the fact that you used the series expansion centered at
, not 
, and so the sign on each derivative at is opposite of what it should be. I'm sure you can figure out the radius of convergence from here.
- - -
6. Note that this is already a polynomial, so the Taylor series will strongly resemble this and will consist of a finite number of terms. You can get the series by evaluating the derivatives at the given point, or you can simply rewrite the polynomial in
as a polynomial in 
.
If you expand this, you will end up with
again, so the Taylor series must converge everywhere.
I'll outline the second method. The idea is to find coefficients so that the right hand side below matches the original polynomial:
You would expand the right side, match up the coefficients for the same-power terms on the left, then solve the linear system that comes out of that. You would end up with the same result as with the standard derivative method, though perhaps more work than necessary.
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7. It would help to write the square root as a rational power first:
The pattern should be fairly easy to see.
By the ratio test, the series converges if
so that the ROC is 4.
- - -
10. Without going into much detail, you should have as your Taylor polynomial
Taylor's inequality then asserts that the error of approximation on the interval
is given by
where
satisfies 
on the interval.
We know that
is bounded between -1 and 1, so we know 
will suffice. Over the given interval, we have 
, so the remainder will be bounded above by
which is to say, over the interval, the fourth degree Taylor polynomial approximates the value of 
near 
to within 0.000328.
Clearly, each even-order derivative will vanish, and the terms that remain will alternate in sign, so the Taylor series is given by
Your answer is off by a sign - the source of this error is the fact that you used the series expansion centered at
- - -
6. Note that this is already a polynomial, so the Taylor series will strongly resemble this and will consist of a finite number of terms. You can get the series by evaluating the derivatives at the given point, or you can simply rewrite the polynomial in
If you expand this, you will end up with
I'll outline the second method. The idea is to find coefficients so that the right hand side below matches the original polynomial:
You would expand the right side, match up the coefficients for the same-power terms on the left, then solve the linear system that comes out of that. You would end up with the same result as with the standard derivative method, though perhaps more work than necessary.
- - -
7. It would help to write the square root as a rational power first:
The pattern should be fairly easy to see.
By the ratio test, the series converges if
so that the ROC is 4.
- - -
10. Without going into much detail, you should have as your Taylor polynomial
Taylor's inequality then asserts that the error of approximation on the interval
where
We know that
which is to say, over the interval
X is in the second quadrant means that x/2 is in the first quadrant.
Consider the right triangle drawn in the figure. Let tan(x/2)=a.
Then, let the length of the opposite side to x/2 be a, the adjacent side be 1 and the hypotenuse be square root of a squared +1, as shown in the figure.
sin(x/2)=|opp side|/ |hypotenuse| =
cos (x/2) = |adj side|/ |hypotenuse| =
from the famous identity: sin(2a)=2sin(a)cos(a), we have:
2sin(x/2)cos (x/2)=sin(x)
thus
(3a-1)(a-3)=0
thus a=1/3 or a=3
thus tan(x/2)=1/3 or tan(x/2)=3
Answer: {1/3, 3}
Consider the right triangle drawn in the figure. Let tan(x/2)=a.
Then, let the length of the opposite side to x/2 be a, the adjacent side be 1 and the hypotenuse be square root of a squared +1, as shown in the figure.
sin(x/2)=|opp side|/ |hypotenuse| =
cos (x/2) = |adj side|/ |hypotenuse| =
from the famous identity: sin(2a)=2sin(a)cos(a), we have:
2sin(x/2)cos (x/2)=sin(x)
thus
(3a-1)(a-3)=0
thus a=1/3 or a=3
thus tan(x/2)=1/3 or tan(x/2)=3
Answer: {1/3, 3}