The Bernoulli polynomials are defined by
- Find explicit formulas for the first Bernoulli polynomials in the cases .
- Use mathematical induction to prove that is a degree polynomial in , where the degree term is .
- For prove that .
- For prove that
- Prove that
for .
- Prove that for ,
- Prove that for ,
- We start with the initial condition . This gives us
Now, using the integral condition to find ,
Thus,
Next, using this expression for we have
Using the integral condition to find ,
Thus,
Next, using this expression for we have
Using the integral condition to find ,
Thus,
Next, using this expression for we have
Using the integral condition to find ,
Thus,
Finally, using this expression for we have
Using the integral condition to find ,
Thus,
- Proof. We have shown in part (a) that this statement is true for . Assume then that the statement is true for some positive integer , i.e.,
Then, by the definition of the Bernoulli polynomials we have,
where for . Then, taking the integral of this expression
Hence, the statement is true for the case ; hence, for all positive integers
- Proof. From the integral property in the definition of the Bernoulli polynomials we know for ,
Then, using the first part of the definition we have ; therefore,
Thus, we indeed have
- Proof. The proof is by induction. For the case we have
Therefore,
Since , the stated difference equation holds for . Assume then that the statement holds for some positive integer . Then by the fundamental theorem of calculus, we have
Therefore,
Hence, the statement is true for the case , and so it is true for all positive integers
- Proof. (Let’s assume Apostol means for to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,
Now, we want to express the numerator as a telescoping sum and use part (d),
Thus, we indeed have
- Proof.
Incomplete. I’ll try to fix parts (f) and (g) soon(ish).
I tried induction for (f):
For n=1 you have
and so the equality holds. Now suppose it holds for
, and we want to show that
We observe that
and that
and so the induction hypothesis gives us that
Thus,
and
are primitives of the same function, and differ only by a constant. So to prove that they are equal, it suffices to exhibit one point of equality. \\
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If is even, then we have
by part (c), so we’re done. \\
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If is odd, we wish to find a point at which
By the definition, and making a substitution, we have
Now I’ll refer back to an earlier exercise, exercise 7 from section 3.20, which tells us that if is non-negative and has zero integral over some interval, then it must be zero at every point on the interval where it’s continuous (the same holds if is non-positive). Since
is continuous everywhere, we may apply the result here. There are two cases:
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)
does not change sign on . Then by the above result, it must be zero everywhere, and we may exhibit any point on to complete the proof.
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)
changes sign on . Then by the intermediate value theorem, it vanishes at some
. In either case, we have found a point of equality.
oh no something went wrong, here’s the latex:
I tried induction for (f):
For n=1 you have
\[P_1(1-x)=(1-x)-\frac{1}{2}=-x+\frac{1}{2}=-(x-\frac{1}{2})\]
and so the equality holds. Now suppose it holds for $n=m\ge1$, and we want to show that
\[P_{m+1}(1-x)=(-1)^{m+1}P_{m+1}(x)\]
We observe that
\[\frac{d}{dx}\bigg(P_{m+1}(1-x)\bigg)=-P_{m+1}'(1-x)=-(m+1)P_m(1-x)\]
and that
\[\frac{d}{dx}\bigg((-1)^{m+1}P_{m+1}(x)\bigg)=(-1)^{m+1}(m+1)P_m(x)\]
and so the induction hypothesis gives us that
\[\frac{d}{dx}((-1)^{m+1}P_{m+1}(x))=\frac{d}{dx}(P_{m+1}(1-x))\]
Thus, $P_{m+1}(1-x)$ and $(-1)^{m+1}P_{m+1}(x)$ are primitives of the same function, and differ only by a constant. So to prove that they are equal, it suffices to exhibit one point of equality. \\
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If $m+1$ is even, then we have $P_{m+1}(1-0)=P_{m+1}(0)$ by part (c), so we’re done. \\
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If $m+1$ is odd, we wish to find a point $x$ at which
\[P_{m+1}(1-x)=-P_{m+1}(x) \iff P_{m+1}(1-x)+P_{m+1}(x)=0\]
By the definition, and making a substitution, we have
\[\int_{0}^{1}\bigg(P_{m+1}(1-x)+P_{m+1}(x)\bigg)dx=0\]
Now I’ll refer back to an earlier exercise, exercise 7 from section 3.20, which tells us that if $f$ is non-negative and has zero integral over some interval, then it must be zero at every point on the interval where it’s continuous (the same holds if $f$ is non-positive). Since $P_{m+1}$ is continuous everywhere, we may apply the result here. There are two cases:
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$i$) $P_{m+1}(1-x)+P_{m+1}(x)$ does not change sign on $[0,1]$. Then by the above result, it must be zero everywhere, and we may exhibit any point on $[0,1]$ to complete the proof.
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$ii$) $P_{m+1}(1-x)+P_{m+1}(x)$ changes sign on $[0,1]$. Then by the intermediate value theorem, it vanishes at some $c\in[0,1]$. In either case, we have found a point of equality.
f) can be proved by induction as well (e.g. similar to d) part).
Actually, while it is probably doable to do f) by induction (e.g. to prove that later odd Bernoulli numbers are 0), the other suggested solution here looks more appropriate.
The proof of part (d) is wrong since he assumes that p_m+1(0)=0 which is clearly not true (in general) from part a
In part d, he assumes that P_m+1(0)=0 which is clearly wrong from part a, but he got lucky since it got cancelled when he took the difference P_m+1(x+1) – P_m+1(x)
For part (f), I found a way to do it online. First, prove that if some polynomials {Qn} satisfy the three conditions above then the polynomial Qn(x) must equal Pn(x) for every n. Then define polynomial Qn(x) := (-1)^n Pn(1-x) and show that this polynomial satisfies the three conditions above. Part (g) follows from part (f) and part (c).
Hello, I think you have to add a constant after the indefinite integrals in (d). Thanks for your solutions!
Solution of letter f is in this link, page 8: https://www.mathi.uni-heidelberg.de/~theiders/PS-Analysis/Bernoullische_Polynome.pdf
With letter f proved, it’s trivial solve the part g.