Four of the previous exercise (here, here, here, and here, Section 10.9, Exercises #11 – #14) suggest a formula
In the formula indicates a polynomial of degree , with the term of lowest degree being and the highest degree term being (we call a polynomial of degree with highest term a monic polynomial of degree ). Use mathematical induction to prove this formula, without justifying the manipulations with the series.
Proof. For the case , we have
from the first exercise linked above (Section 10.7, Exercise #11). Letting , a degree polynomial, with term of lowest degree being and term of highest degree being . Therefore, the statement holds for the case . Assume then that the statement is true for some , so
Differentiating both sides (assuming without justification that we can do this) we obtain
where is a polynomial of degree with the term of lowest degree being and that of the highest degree being . Hence, if the statement is true for then it is true for ; thus, the statement is true for all positive integers
Let be a function differentiable everywhere such that
- Prove that and conjecture and prove a similar formula for .
- Conjecture and prove a formula for in terms of for all positive integers .
- Prove that and compute
- Prove that there exists a constant such that
for all . Find the value of the constant .
- Proof. We can compute this using the functional equation:
Next, we conjecture
Proof. Again, we compute using the functional equation, and the above formula for ,
- We conjecture
Proof. The proof is by induction. We have already established the cases and (and the case is the trivial ). Assume then that the formula holds for some integer . Then we have
Thus, if the formula holds for , it also holds for . Hence, by induction it holds for all integers
- Proof. Using the functional equation
Then, since the derivative exists for all (by hypothesis) we know it must exist in particular at . Using the limit definition of derivative, and the facts that and we have
- Proof. Since the derivative must exist for all (by hypothesis) we know that the limit
must exist for all . Using the functional equation for we have
But then, from part (c) we know and from this exercise (Section 6.17, Exercise #38) we know
Therefore, we have
If is a product of functions prove that the derivatives of are given by the formula
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 .
Proof. The proof is by induction. For we have
These equalities follow from the co-relations of sine and cosine (Theorem 2.3 part (d) on page 96 of Apostol). Thus, the formulas are true for the case . Assume then that they are true for some . For we then have
Similarly, for we have
Therefore, the theorem follows by induction for all positive integers
Assume is a polynomial with . Define
Compute the values of .
Assume is a non-negative integer (otherwise is undefined at ). Then, we make the following claim:
Claim: The polynomial has derivatives at 0 given by the following
Proof. Since is a polynomial we may write,
Furthermore, since is given we know . Now, multiplying by we have
Next, we will use induction to prove that the th derivative of for is given by
for constants . Since the derivative is given by
for constants , we see that the formula holds for . Assume then that it holds for some ,
Then, taking the derivative of this we have,
Hence, the formula holds for all . But then, if we have
If then for all ; hence,
be the product of functions with derivatives . Find a formula for the derivative of , and prove that it is correct by mathematical induction.
Furthermore, show that
for those at which for any .
Claim: If , then
Proof. For the case , from the usual product rule we have
Thus, our claimed formula holds in this case. Assume then that it is true for some integer . Then, if we have,
Therefore, if the formula holds for then it also holds for ; thus, it holds for all positive integers
Next, we prove the formula for the quotient, .
Proof. Let be defined as above, and let be a point at which for any . Then,
Consider a polynomial of degree ,
- Prove that if and then there is an degree polynomial such that .
- Prove that is a polynomial of degree for any .
- Prove that if and , then , for a polynomial of degree .
- If has real zeros (i.e., there exist numbers such that ), then for all , and for all .
is a polynomial of degree , and if
for distinct , then
where is then an degree polynomial
Proof. We compute, using the binomial theorem,
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
Proof. From part (b) we know that if is a polynomial of degree , then is also a polynomial of the same degree. Further,
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
Proof. The proof is by induction. First, we consider the case in which . Then . Since the assumption is that there are distinct real such that , we know there exist such that
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,
for all . Further, since for , and by assumption for , we have for . But then,
means is a polynomial of degree as well