Historical Context of the Bernoulli Family
Although many people introduced to the name ”Bernoulli” might assume that it is just one mathematician, the truth of the matter is that this was an entire family of mathematicians. This family, although originally from Antwerp, moved to Basel, Switzerland later on in their life. A family consisting of more than eight outstanding mathematicians, the Bernoulli’s lived throughout the course of three generations and most of them made enormous contributions in the areas of mathematics and physics that mathematicians and physicists still use today. Of these people, the most notable three of them are Jacob Bernoulli, Johann Bernoulli, and Daniel Bernoulli. Johann Bernoulli made many major contributions to infinitesimal calculus and educated Leonhard Euler, one of the greatest mathematicians of all time. Daniel Bernoulli, the family member who worked in physics the most out of the three, is mostly remembered for his connection of mathematics and mechanics, a branch of physics. One of his most important works is the Bernoulli Principle, which is a specific example of the Conservation of Energy Law. Lastly, Jacob Bernoulli, arguably the member of the Bernoulli family who did the most contributions of mathematics, made these contributions in many different areas of mathematics. This family left a legacy unique than that of many mathematicians in that most mathematicians were prodigies whose mathematical abilities surpassed that of the rest of the family, but multiple members of the Bernoulli family proved their expertise in different areas of mathematics, therefore making them one of the most elite mathematical families of all time.
The Life of Jacob Bernoulli As previously stated, Jacob Bernoulli was the most prominent of all the mathematicians of the Bernoulli family. His first name can be either written as Jacob or as Jakob. He was born in Basel, Switzerland, and chose to study theology due to the demands of his father. He also studied, however, mathematics and astronomy, even though his parents strongly discouraged him to take part in any kind of mathematics. Jacob did not listen; from 1676 to 1682, he traveled around Europe in order to study the recent discoveries of leading physicists and mathematicians at the time, such as Johannes Hudde, Robert Hooke, and Robert Boyle. Interested more in physics than mathematics at this time, Jacob moved back to Switzerland and started to teach mechanical physics at the University of Basel in 1683. Around the same time, Jacob married Judith Stupanus and together they had two children. He began to do more research around this time as well, and had several meetings with many other leading mathematicians. Due to this research, Jacob switched from mechanics to mathematics and became professor of mathematics at the University of Basel in 1687, a position he had until the end of his life. At the start of his career as a mathematics professor, he started to tutor his brother Johann Bernoulli on mathematical topics, specifically calculus. Together they studied Leibniz’s paper on differential calculus, and this is how Johann obtained his love for infinitesimal calculus. Unfortunately, the friendly bond the two brothers had was quickly broken as Johann started to become increasingly better at mathematics. By 1697, the relationship was over between the two and they had become bitter rivals. A couple of years later, in 1705, Jacob Bernoulli died. To no surprise, however, Jacob Bernoulli wanted his mathematics to continue even after his death. He chose a logarithmic spiral to be his gravestone, but the stonemasons constructed an Archimedean spiral instead. Before his death, Jacob explained that the logarithmic spiral ”may be used as a symbol, either of fortitude and constancy in adversity, or of the human body, which after all its changes, even after death, will be restored to its exact and perfect self.”
Jacob Bernoulli’s Mathematical Works
Jacob’s first major important contribution to mathematics was a pamphlet on the parallels of logic and algebra that was published in 1685. This was followed by work on probability during the same year, and two years later, Jacob published works on geometry. This extremely important work led to the conclusion that it is possible to divide any triangle into four equivalent parts by using two perpendicular lines. It is said that in 1683 Bernoulli discovered the constant e by studying topics related to compound interest by finding the value of
2 the limit as n tends to infinity of a certain mathematical expression. Although the topic of compound interest was not as interesting as the other areas of mathematics he was involved with, the number e is used in many applications of mathematics today. In 1689 Jacob started to publish works on infinite series, and although a mathematician by the name of Mengoli discovered that the infinite series 1/n diverges 40 years earlier, Jacob believed that he had discovered the result through his work. Bernoulli was also able to show that 1/n squared converged to a finite limit less than 2 as n tended to infinity. Many years later, Leonhard Euler, a pupil of Johann Bernoulli, was able to find the sum of that series. Jacob also started to work on the exponential series after his discovery of the number e. In 1960 Jacob made a huge contribution in the area of differential equations. He discovered that determining the isochrone is the same thing as solving a first-order differential equation, and the isochrone is equivalent to a curve of constant descent, and it is the trajectory that a particle will take to descend because of gravity no matter where the starting point is. Bernoulli solved this differential equation in 1696, and it is now called the Bernoulli differential equation. This had a major impact because a differential equation could be shaped into the format of the Bernoulli differential equation, and therefore the differential equation then becomes trivial to solve after the transformation. Jacob also studied the curves of the parabola, epicycloids, and the logarithmic spiral around 1692. Eight years after his death, in 1713, Jacob’s Ars Conjectandi was published. This was arguably Jacob’s most original piece of work. This work has very significant ideas in probability theory. In this work, Jacob discussed the works of other mathematicians in the field of probability, such as Leibniz, Prestet, and van Schooten. Aside from probability, Jacob also explains a very prominent concept in mathematics that helps describe many infinite power series: the Bernoulli Numbers.
Bernoulli Numbers and Their Applications
Although Jacob’s work Ars Conjectandi focuses mainly on probability, he also talks about the Bernoulli numbers. These are considered by many to be very mysterious, and are found in many various places of mathematics, such as number theory, analysis, and differential topology. These numbers help greatly in the Euler-Maclaurin summation formula, in which the numbers help mathematicians compute series that converge slowly. These numbers appear in number theory, and more specifically Fermat’s Theory, and the first recorded people to use these numbers include Abraham De Moivre and Leonhard Euler. Another important application of the Bernoulli Numbers is that they help in showing the series expansion in trigonometric and hyperbolic functions, and the proof portrayed below will show how the Bernoulli Numbers can be used to find the series expansion for tan(x).
3 The Generating Function for Bernoulli Numbers
In order to completely understand the process to determine the power series for tan(x), a person must first understand how the Bernoulli generating function came to be. The math that is used to show the generation function will not be covered in great detail, and therefore this is more of a simplified version whose purpose is to show the main methods used. The process started with Pierre de Fermat, who wanted to compute the sums for the following form:
n X p sp(n) = k k=1 From here, the mathematician Johann Faulhaber found simplified formulas for these sums for p less than 18. Some examples of these are as follows:
s0(n) = n
n(n − 1) s (n) = 1 2
n(n − 1)(2n − 1) s (n) = 2 6
n2(n − 1)2 s (n) = 3 4
n(n − 1)(2n − 1)(3n2 − 3n − 1) s (n) = 4 30
n2(2n2 − 2n − 1)(n − 1)2 s (n) = 5 12 This is the information that Jacob Bernoulli had when starting his work on Bernoulli Numbers. From some of Haulhaber’s formulas, Bernoulli discovered that these polynomials had a specific pattern. They followed the form
1 1 p s (n) = np+1 − np + np−1 + 0np−2 + ... p p + 1 2 12 He found that the coefficients of each term are independent of p and more specifically these numbers are the Bernoulli Numbers. By this he found the formula:
p X Bk p! np+1−k k! (p + 1 − k)! k=0
Where Bk are the Bernoulli Numbers
Ironically, the next person to continue in finding this generating equation was Leonhard Euler, the pupil of Johann Bernoulli, the major rival of Jacob Bernoulli. Euler decided to try to form a Taylor series expansion where the derivative part of the function would be the Bernoulli Numbers.
4 ∞ ∞ X xk X xk f(x) = f (k)(0) f(x) = B k! k k! k=0 k=0 From here, Euler used the Cauchy product method by multiplying the infinite series for f(x) with the expo- nential infinite series. The Cauchy product is a method for convoluting infinite series together.
∞ ! ∞ ! X xk X xk f(x)ex = B k k! k! k=0 k=0
∞ k ! X X xi xk−i f(x)ex = B i i! (k − i)! k=0 i=0
∞ k ! X X Bi f(x)ex = xk i!(k − i)! k=0 i=0
∞ k ! X X k xk f(x)ex = B i i k! k=0 i=0
If we then let ck be denoted by:
k k−1 X k X k c := B = B + B = k + B k i i i i k k i=0 i=0
th The previous step can be done because the k term in the series is simply Bk and the rest can by denoted x just as k, giving the explanation as to why ck = k + Bk. From here what is left to do is to find f(x)e in a more simplified form:
∞ ∞ X xk X xk f(x)ex = (k + B = + f(x) = xex + f(x) k k! (k − 1)! k=0 k=1 And in conclusion this shows that the generating function that Euler was looking for was:
xex f(x) = ex − 1 And when specified that ex > 0 for all x, then f(x) can be rewritten as:
x f(x) = 1 − e−x With some algebraic manipulation, this can finally be rewritten as the popular form of the Bernoulli generating function x f(x) = ex − 1
Proof for the Series Expansion of Tan(x) Using Bernoulli Numbers
5 Claim: The series expansion for tan(x) using Bernoulli Numbers is:
∞ k k k 2k−1 X B2k(−1) (4) (4 − 1)(x) 2k! k=0 In order to prove this, one must know these background identities:
Euler’s Formula: Euler’s formula states that:
eix = cos(x) + isin(x) and by replacing i with -i, we can state that
e−ix = cos(x) − isin(x)
If one adds these two equations together, one gets that
eix + e−ix = 2cos(x) or in other words,
eix + e−ix cos(x) = 2 By subtracting the second equation from the first equation, one can state that
eix − e−ix = 2isin(x) or in other words,
eix − e−ix sin(x) = 2i These two equations are handy in that they will help in the last part of the proof.
Trigonometric Identity: In order to finish off the proof, one must know that
tan(x) = cot(x) − 2cot(2x)
Proof: Now that one has acknowledged the background information, one can now start the formal proof. One begins with the generating function for Bernoulli Numbers:
∞ k X Bk(t) t = k! et − 1 k=0 One will then proceed by just using the even powers of this series, and therefore the following summation is created so that all the odd powers are 0.
∞ k ∞ k X Bk(t) X Bk(−t) t −t + = + k! k! et − 1 e−t − 1 k=0 k=0 By adding the left side of the equation together and joining the right side of the equation into a single term,
6 we can simplify the equation into:
∞ 2k −t t X B2k(t) t(e − 1) − t(e − 1) 2 = 2k! (et − 1)(e−t − 1) k=0 The numerator of the right side of the equation can be simplified a bit further, and one gets:
∞ 2k −t t X B2k(t) t(e − e ) 2 = 2k! (et − 1)(e−t − 1) k=0 Before one can continue, further algebraic manipulation needs to be done to the right side of the equation in order to simplify it down further. From the numerator of the right side of the equation one will need to split the quantity inside of the parenthesis as the product of two conjugates.
t(e−t − et) = t(e−t/2 + et/2)(e−t/2 − et/2)
From the denominator of the right side, some factoring needs to be done from both the first term and the second term
et − 1 = et/2(et/2 − e−t/2) and
e−t − 1 = e−t/2(e−t/2 − et/2)
Therefore the equation on the right side can be finally written as
t(e−t/2 + et/2)(e−t/2 − et/2) et/2(et/2 − e−t/2)e−t/2(e−t/2 − et/2) After final cancellation, the right side of the equation can finally be written as
t(e−t/2 + et/2) et/2 − e−t/2 The complete simplified equation is now
∞ 2k −t/2 t/2 X B2k(t) t(e + e ) 2 = 2k! et/2 − e−t/2 k=0 After the equation has been established, one needs to replace t with 2ix
∞ 2k −(2ix)/2 (2ix)/2 X B2k(2ix) (2ix)(e + e ) 2 = 2k! e(2ix)/2 − e−(2ix)/2 k=0 With simple simplification on both sides, one can see that
∞ k k 2k ix −ix X B2k(−1) (4) x 2ix(e + e ) 2 = 2k! eix − e−ix k=0
7 The x from the right side of the equation needs to be moved to the left side of the equation, and this is not a problem since the summation depends on k
∞ k k 2k−1 ix −ix X B2k(−1) (4) x 2i(e + e ) 2 = 2k! eix − e−ix k=0 One then needs to cancel out the 2 on both sides to get
∞ k k 2k−1 ix −ix X B2k(−1) (4) x i(e + e ) = 2k! eix − e−ix k=0 From this moment, one needs to recall Euler’s formula given in the background section just before the proof. The right side of the equation can be rewritten as
i(eix + e−ix) (eix + e−ix) = eix − e−ix eix−e−ix i From Euler’s Formula one can see that
2cos(x) = eix + e−ix and
eix − e−ix 2sin(x) = i From here it is easy to see that
∞ k k 2k−1 X B2k(−1) (4) x 2cos(x) = 2k! 2sin(x) k=0 And finally by cancelling out the 2 on the right side of the equation and with a very simple trigonometric identity one can finally see that
∞ k k 2k−1 X B2k(−1) (4) x = cot(x) 2k! k=0 A common mistake is to think that in order to find the power series for tan(x), one simply has to take the reciprocal of the quantity inside the sigma notation. The answer, however, is not that simple. The final concept that needs to be recalled is the trigonometric identity
tan(x) = cot(x) − 2cot(2x)
The focus will now be put on the right side of the equation. The power series for cot(x) will be substituted into cot(x). The right side of the equation will turn into
∞ k k 2k−1 ∞ k k 2k−1 X B2k(−1) (4) x X B2k(−1) (4) (2x) − 2 2k! 2k! k=0 k=0 The two on the right sigma notation will be placed inside and multiplied into the term, so that
8 ∞ k k 2k−1 ∞ k k 2k 2k−1 X B2k(−1) (4) (2x) X B2k(−1) (4) (2) (x) −2 = − 2k! 2k! k=0 k=0 Once again, more simplification can be done so that
∞ k k 2k 2k−1 ∞ k 2k 2k−1 X B2k(−1) (4) (2) (x) X B2k(−1) (4) (x) − = − 2k! 2k! k=0 k=0 From here the right side of the equation is finally
∞ k k 2k−1 ∞ k 2k 2k−1 X B2k(−1) (4) x X B2k(−1) (4) (x) − 2k! 2k! k=0 k=0 Both summations can now be combined into one single summation, and one will finally get that
∞ k k k 2k−1 X B2k(−1) (4) (1 − 4 )x tan(x) = 2k! k=0 The next step is to manipulate the quantity algebraically so that the result looks like the most common used form of the power series for tan(x)
∞ k k k 2k−1 ∞ k 2k 2k 2k−1 X B2k(−1) (4) (1 − 4 )x X B2k(−1) (2) (1 − 2 )x = 2k! 2k! k=0 k=0 Although this answer is technically correct, many mathematicians write the series expansion by starting k at 1, and not 0, because the first term by using k=0 would put x in the denominator, and the tangent of x would not be defined. Therefore an alternate solution can be
∞ k−1 2k 2k 2k−1 X B2k(−1) (2) (2 − 1)(x) 2k! k=1
References
1. http://mathworld.wolfram.com/BernoulliNumber.html
2. http://home.iitk.ac.in/~tmk/reachout/BernoulliNum.pdf 3. https://wj32.org/wp/2011/10/30/power-series-of-tanx-cotx-cscx/
4. https://en.wikipedia.org/wiki/Bernoulli_family
5. http://www.maths.tcd.ie/pub/HistMath/People/Bernoullis/RouseBall/RB_Bernoullis.html
6. http://www.storyofmathematics.com/18th_bernoulli.html
7. https://en.wikipedia.org/wiki/Jacob_Bernoulli
8. http://numbers.computation.free.fr/Constants/Miscellaneous/bernoulli.html
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