Integrate the binomial series for to obtain the power-series expansion
Incomplete.
Integrate the binomial series for to obtain the power-series expansion
Incomplete.
Show that
where
Since we do not have the binomial theorem for real powers (we have proved a formula for for integers , but in this case the power we have is a real number ), we use induction to determine the th derivative of the function . First, we compute a few derivatives of ,
So, we conjecture
We have shown this is true for , and if we suppose it is true for a positive integer then we have
Therefore, the formula holds for all positive integers. Now we can compute the Taylor polynomial directly,
Prove the identity:
Proof. Using the hint (that ) we start with the expression on the left,
(The interchange of the sum and integral is fine since it is a finite sum. Those planning to take analysis should note that this cannot always be done in the case of infinite sums.) Now, we have a reasonable integral, but we still want to get everything back into the form of the sum on the right so we make the substitution , . This gives us new limits of integration from 1 to 0. Therefore, we have
If is a product of functions prove that the derivatives of are given by the formula
where
is the binomial coefficient. (See the first four exercises of Section I.4.10 on page 44 of Apostol, and in particular see Exercise #4, in which we prove the binomial theorem.)
Proof. The proof is by induction. Letting and we use the product rule for derivatives
So, the formula is true for the case . Assume then that it is true for . Then we consider the st derivative :
Here, we use linearity of the derivative to differentiate term by term over this finite sum. This property was established in Theorem 4.1 (i) and the comments following the theorem on page 164 of Apostol. Continuing where we left off we apply the product rule,
where we’ve reindexed the first sum to run from to instead of from to . Then, we pull out the term from the first sum and the term from the second,
Now, we recall the law of Pascal’s triangle (which we proved in a previous exercise) which establishes that
Therefore, we have
Hence, the formula holds for if it holds . Therefore, we have established that it holds for all positive integers .
Consider the function for a positive integer . Prove that
Conclude that by considering the limit of this as .
Proof. We recall the binomial theorem,
So, then we calculate,
Taking the limit as of both sides of the equation we conclude,
Transform the given expression to the canonical form to prove that they are, in fact, polynomials.
Consider a polynomial of degree ,
is a polynomial of degree , and if
for distinct , then
Thus,
where is then an degree polynomial
Now, we want to group together the coefficients on the like-powers of . So we pull out the terms corresponding to , , etc. This gives us,
In the final line, we rewrote the coefficients as sums to view them a little more concisely. Either way, since and all of the ‘s are constants, we have is some constant for each , say and we have,
Hence, is a polynomial of degree
So, by part (a), we have
where is a polynomial of degree . Thus,
But, if is a polynomial of degree , then by part (b) again, so is . Hence,
for a degree polynomial, as requested
Thus,
Hence, the statement is true for . Assume that it is true for some . Then, let be a polynomial of degree with distinct real zeros, . Then, since , using part (c), we have,
where is a polynomial of degree . Further, we know there are distinct values such that . (Since for and for with since all of the are distinct.) Thus, by the induction hypothesis, every coefficient of is 0 and for all . Thus,
But, since all of the coefficients of are zero, we have for . Hence,
Thus, all of the coefficients of are zero and for all . Hence, the statement is true for the case , and therefore, for all
where for (we have by assumption).
Then, there are distinct real for which . (Since there are distinct real values for which , and so at each of these values .) Thus, by part (d), for and for all . But then,
and
for all . Further, since for , and by assumption for , we have for . But then,
means is a polynomial of degree as well
This is the identity requested
Thus,
since the second term is strictly positive for positive integers. This establishes the inequality on the left.
For the inequality on the right we use part (a) to get,
Thus,
In the second to last line we have just replaced each term in the sum by , giving the inequality. This establishes both sides of the requested inequality
For a positive integer , then ; hence, the inequalities hold in the case . Assume then that they hold for some .
For the left inequality,
This establishes the left inequality for all .
For the right inequality, assume it is true for some , then
Hence, the right inequality is true for all . Therefore, the inequalities requested are indeed true for all
So, letting and we have,
If , then , so using the Binomial theorem we have,
Where we know the inequality is strict since there is at least one term (which is necessarily positive) in since .
Next, we prove the middle inequality,
By part (a) we know,
Further, for we have for all . Thus, we know that,
Hence,
for all . Therefore, we have established the second inequality,
Finally, we prove the right inequality,
Here we expand the first few terms and use a previous result,
for all . In the second to last line we used this result on the th powers of a real number (in this case ). This completes the proof for all of the inequalities requested
Prove the binomial theorem:
Further, prove the formulas:
On the right,
Hence, the formula is true for the case .
Assume then that the formula is true for some . Then we have,
Thus, if the formula is true for the case then it is true for the case . Hence, we have proved the formula for all
As an application of the binomial theorem we then prove the two formulas.
For the first, apply the binomial theorem with . Then,
For the second, apply the binomial theorem with and . Then,