12. Inner Product Spaces

Transcription

1 1. Inner roduct Spaces 1.1. Vector spaces A real vector space is a set of objects that you can do to things ith: you can add to of them together to get another such object, and you can multiply one of them by any real number to get another such object. There s a set of axioms that a vector space must satisfy; you can find these in other textbooks. Similarly, a complex vector space is a set of objects that you can do to things ith: you can add to of them together to get another such object, and you can multiply one of them by any complex number to get another such object. If you are speaking about a real vector space, you call any real number a scalar. If you are speaking about a complex vector space, you call any complex number a scalar. If you have a (finite) list of vectors v 1, v, v 3,..., v n, the most general ay that you can combine these vectors together to get another vector is to choose a list of scalars α 1, α, α 3,..., α n and form the linear combination α 1 v 1 + α v + α 3 v α n v n Examples of real vector spaces include R N, here the vectors are N-tuples of real numbers (most familiarly, R the plane hich is the set of ordered pairs of real numbers and R 3 threedimensional space hich is the set of ordered triples of real numbers). The basic operations look like this: if x = (x 0, x 1,..., x N 1 ), y = (y 0, y 1,..., y N 1 ), and α is any real number, then x + y = (x 0 + y 0, x 1 + y 1,..., x N 1 + y N 1 ) and αx = (αx 0, αx 1,..., αx N 1 ). The zero vector is (0, 0,..., 0).Other examples include many different varieties of vector spaces hose members are real valued functions on a certain domain. When you deal ith such function spaces, you have to be careful of a fe things. One is that each function must have the same domain each function must be defined everyhere (actually, for spaces here membership is determined by some kind of integrability condition, almost everyhere is good enough). As an example, let s consider a set of real-valued functions suitable for feeding into the Laplace transform: the set of locally integrable functions on the interval [0, ) that don t gro too fast at. Is f(t) = ln t a member of this set? Yes, although its domain isn t quite all of [0, ). It blos up at 0, but not so badly as to spoil its integrability there. Is f(t) = 1 t a member of this set? No because e need to kno hat it does beteen 1 and. As such, it ould be silly to ask hat the Laplace transform of this function is. You also must make sure that your set of functions really is a vector space. This primarily means that you have to check that the sum of any to such functions is also such a function, and that a constant times any such function is still such a function. Examples of function spaces include the folloing: Description Symbol Continuous functions on [a, b] C[a, b] eriodic continuous functions C(T ) Continuous functions on the hole real line C(R) Functions ith one continuous derivative on a set S (S can be [a, b], T, R, etc.) C 1 (S) Bounded (actually, essentially bounded ), functions on a set S L (S) Absolutely integrable functions on S L 1 (S) Square integrable functions on S L (S) Examples of complex vector spaces include C N, the set of N-tuples of complex numbers, hich is just like R N, except that both the entries (components) of the vectors and the scalars are alloed 8

2 to be complex numbers, and various spaces of complex valued functions. Be sure to understand that e are talking about having the values of the function being complex e are not assuming that these are functions of a complex variable. As an example, e inx is a complex-valued function of the real variable x. Our list of complex function spaces that e are likely to encounter is exactly the same as the list above, and hen e say exactly, e mean that e are using the exact same symbols to name these spaces. When e use a symbol like C(T ), e are not committing ourselves as to hether e mean the real-valued functions or the complex-valued functions e either have to make that clear in the surrounding context, or else admit that in the particular case e have in mind it doesn t much matter hich one e mean. The study of vector spaces in general falls under the label linear algebra. In that study, sets of vectors (that is, sets of elements of some vector space) are probed as to hat their span is and as to hether or not they are linearly independent (independent for short). For further details, see any reasonably-ritten textbook on the subject. Here are to definitions out of all of that study: if you have a function (or mapping, or map, etc. the same idea has many names) T that has as its domain the things that you can feed this mapping all of one vector space, and that has as its output elements of some other (ell, possibly other no rule says it can t be the same one) vector space, then T is a linear transformation (or linear mapping or linear hatzit or just plain linear) if: T (αu + βv) = αt (u) + βt (v) for all vectors u and v and all scalars α and β. (1.1) A linear functional is something hose domain is some vector space hich takes vectors as input that gives scalars (real numbers if it is a real vector space and complex numbers if it is a complex vector space) as output, and that satisfies equation (1.1). By no you should be familiar ith the principle that the thing that takes in functions and gives back Fourier coefficients is, by any reasonable definition, linear. 1.. Real inner product spaces The vector space R N comes equipped ith a very special geometrically-inspired algebraic operation called the dot product. The dot product takes in to vectors and gives you back a real number a scalar. Synonyms for the ords dot product include inner product and scalar product. We ill rite the dot product of the to vectors u and v in at least to different notations: as u v or as u, v. The dot product on R N is defined as follos: if x = (x 0, x 1,..., x N 1 ) and y = (y 0, y 1,..., y N 1 ), then x y = x, y = x 0 y 0 + x 1 y x N 1 y N 1 = N 1 k=0 x k y k (1.) The geometric description of the dot product is the folloing: if θ is the angle beteen x and y, then ( ) x y θ = cos 1 (1.3) x y here x = x = x 0 + x x N 1 = x x (1.4) is the length or norm of the vector x. If the dot product of to vectors is zero, then the angle beteen them is a right angle and e call the to vectors orthogonal. We also notice in (1.4) that computations of lengths can use the dot product, as there seems to be a close relationship. (There 83

3 is also the familiar idea that in any problem involving lengths and the ythagorean Theorem, it is frequently easier to ork ith the square of a distance than it is to ork ith the distance itself.) What if e have a real vector space that is not R N such as one of the function spaces? We may, under some circumstances, be able to define something called an inner product on that vector space. An inner product on a real vector space is something that has all of the vital properties of the dot product on R N if e can figure out hat those properties are. The accepted line of jargon for a real inner product is that it is a positive-definite symmetric bilinear form. What does this mean? It means that it is an object that takes as input to elements of the vector space and gives back a real number - for vectors u and v, let us rite the inner product as u, v. The other ords in this description have the folloing meanings: Symmetric: Bilinear: ositive definite: u, v = v, u for all vectors u and v (1.5) αu + βv, = α u, + β v, and u, αv + β = α u, v + β u, for all u, v,, α and β (1.6) u, u 0 alays and u, u > 0 for u 0 (1.7) The best example that e ill have of something that satisfies all of these properties is to take a space hose elements are functions and to let f, g = fg, or something very much like that. For instance, for functions periodic of period, let us define the standard real inner product of fand g as: f, g = 1 0 f(x)g(x) dx (1.8) It is not hard to sho that this satisfies properties (1.5), (1.6), and (1.7) - except possibly for a little fudging on the second part of (1.7), but let s not orry too much about that no. Any real vector space on hich a real inner product has been defined is a real inner product space Complex inner product spaces We d like to do this for complex vector spaces - but e realize that there are going to have to be some subtle modifications of the detail. In particular, if e ant something positive definite something like (1.7) e are going to have to use something ith a lot of complex conjugates and absolute values in it. Conventional isdom has settled on just hat e need, and it is the folloing: a complex inner product is something that takes as input to vectors from a complex vector space and gives as output a complex number, and it is a positive-definite Hermitian sesquilinear form. Actually, sesquilinear is a part of Hermitian, but I thre it in to make the phrase sound more impressive. Sesquilinear means linear in one factor and conjugate-linear in the other factor - that s something like linear, but ith some stray complex conjugates hanging around. It makes no earthly difference hich factor is hich, but e have to come to some choice and stick to it. As accident ould have it, mathematicians have fallen into the rut of alays putting the complex conjugate on the second factor, hile physicists have fallen into the rut of putting the complex conjugate on the first factor. In hat follos, e ill follo the mathematician s convention - e suppose that a physicist ill just have to read it in a mirror. Here s hat these ords mean: 84

4 Hermitian: Sesquilinear: ositive definite: u, v = u, v for all vectors u and v (1.9) αu + βv, = α u, + β v, and u, αv + β = α u, v + β u, for all u, v,, α and β (1.10) u, u 0 alays and u, u > 0 for u 0 (1.11) We ill give to examples: the standard inner product on C N is defined as follos: if z = (z 0, z 1,..., z N 1 ) and = ( 0, 1,..., N 1 ), then z, = z z z N 1 N 1 = N 1 On a function space the inner product of f and g ill be f, g = k=0 z k k (1.1) to functions periodic of period, the standard complex inner product of f and g is: f, g = 1 0 fg, or to make it specific f(x)g(x) dx (1.13) Any complex vector space on hich a complex inner product as been defined is called a complex inner product space The real Cauchy-Buniakovski-Scharz inequality Theorem the Cauchy-Scharz inequality: in any real inner product space, for any to vectors u and v, u, v u, u v, v (1.14) ith equality holding if and only if one of these vectors is a scalar multiple of the other. The proof uses nothing but the properties of an inner product that is, (1.5), (1.6), and (1.7) and the technique of completing the square. If v = 0, there is nothing to prove, both sides of the inequality being zero, so e assume that v 0. Consider the vector u + λv for any real number λ. By property (1.7), the inner product of this vector ith itself is alays greater or equal to zero (ith equality only if it is the zero vector). Thus: 0 u λv, u λv = u, u u, λv λv, u + λv, λv by (1.6) = u, u λ u, v + λ v, v by (1.5) and (1.6) Since e kno that v, v is positive (by (1.7) again), e can divide both sides of this inequality by it, getting: λ u, v u, u λ + v, v v, v 0 λ u, v v, v 85 λ u, u v, v

5 No complete the square that is, use formula (11.1) from the previous chapter: [ ] λ u, v u, v [ ] u, v v, v λ + u, u v, v v, v v, v ( ) u, v λ u, v u, u v, v v, v v, v This is true for all real λ. If e let λ be that value hich minimizes the left hand side that is, u, v if e let λ be, then the left hand side is zero and the right hand side must be less than or v, v equal to zero. Hence, 0 u, v u, u v, v v, v or u, v u, u v, v As it turns out, there is another notation in hich this fact is usually expressed. From property (1.7), the inner product of a vector ith itself is positive and hence has a real, positive square root. We call this square root the norm of the vector - by analogy to (1.4). That is, in any real inner product space e define the norm of a vector u to be u = u, u (1.15) Using this terminology, e take the square root of both sides of the inequality (1.14) to get: The Cauchy-Scharz inequality (norm version): u, v u v (1.16) u, v One side effect of this is that e can form the fraction and be assured that it is beteen u v 1 and 1. Hence it has an arccosine, and e call that arccosine the angle beteen the to vectors The complex Cauchy-Buniakovski-Scharz inequality Theorem the Cauchy-Scharz inequality: in any real inner product space, for any to vectors u and v, u, v u, u v, v (1.17) ith equality holding if and only if one of these vectors is a scalar multiple of the other. What s the difference beteen this and equation (1.14)? To things: the fact that the vectors involved belong to a complex rather than a real vector space, and the ay that e had to make it the square of the absolute value of something on the left hand side e can t put just the square of a complex number into an inequality and have it be meaningful. The proof uses nothing but the properties of an inner product that is, (1.9), (1.10), and (1.11) and the technique of completing the square for a complex variable. If v = 0, there is nothing to prove, both sides of the inequality being zero, so e assume that v 0. Consider the vector u + λv for any complex number λ. By property (1.1), the inner product of this vector ith itself is alays greater or equal to zero (ith equality only if it is the zero vector). Thus: 0 u λv, u λv 86

6 = u, u u, λv λv, u + λv, λv by (1.6) = u, u λ u, v λ v, u + λλ v, v = u, u λ u, v λ u, v + λλ v, v by (1.9) and (1.10) Since e kno that v, v is positive (by (1.11) again), e can divide both sides of this inequality by it, getting: λ u, v u, v u, u λ λ + v, v v, v v, v 0 Next, employ equation (11.7) of the Completing the Square chapter: λ u, v u, v λ λ v, v v, v u, v u, v + v, v v, v ( λ u, v ) ( ) λ u, v v, v v, v u, v λ v, v + u, v u, v u, u + v, v v, v v, v 0 u, v v, v + u, u v, v 0 u, u v, v u, v v, v 0 This is true for all possible values of λ, especially including that value hich minimizes the left u, v hand side namely, λ =. If e choose that value of λ, then the numerator of the second v, v fraction must be greater or equal to zero - that is, (1.17) must be true. We no repeat the ay e finished off the previous section: e let the norm of the vector be the square root of its inner product ith itself that is, in a complex inner product space, u = u, u (1.18) Given this convention, e take the square root of both sides of (1.17) to get The Cauchy-Scharz inequality (norm version, complex vector space): u, v u v (1.19) 1.6. Orthonormal sets in a real inner product space The driving principle of the vast majority of mathematical ork in inner product spaces is orthogonality. The definition of orthogonal is the same in both real and complex inner product spaces: to vectors are orthogonal if and only if their inner product is zero. With that in mind, e define the concepts of an orthogonal set and an orthonormal set of vectors: Definition: Let {e j } be a set of vectors in a (real or complex) inner product space. The variable j the index to this list of vectors runs through some set of possible values. We are at the moment being deliberately vague as to hether that index set is finite or infinite. Then {e j } is an orthogonal set if and only if: { = 0 if j k e j, e k is (1.0) > 0 if j = k 87

7 The same set is an orthonormal set if and only if: { 0 if j k e j, e k = 1 if j = k (1.1) An orthonormal set is clearly a special case of an orthogonal set. On the other hand, any orthogonal set may be turned into an orthonormal set by merely dividing each element by its on norm. Note also that, although the zero vector is alays orthogonal to everything, e don t ant it as a member of anything that e are illing to call an orthogonal set. Our central problem is this: suppose e have a finite orthonormal set {e j } (no e are specifying that the set of indices the set of possible values for j be a finite set, although e are illing to let it be a very large finite set.) Let v be any vector in the inner product space. Ho closely can e approximate v by a linear combination of the elements of the orthonormal set? More specifically, ho can e choose the coefficients {α j } so that α j e j is as close as possible to v? j But hat do e mean by as close as possible? Surely e mean that the size of the difference is as small as possible. But hat do e mean by size? Well, let s see every inner product space, real or complex, has a built-in notion of size: the norm, as defined in (1.15) and (1.18). That is, our problem is to find α j so that v α j e j is minimized. j To minimize this, it is sufficient to minimize its square. This isn t a ne idea, of course it is used in nearly every calculus problem that asks that a distance be maximized or minimized. The ythagorean theorem just makes orking ith the squares of distances easier than orking ith the distances themselves. So here e go: v j α j e j = v j α j e j, v k α k e j (1.) We used to different names j and k for the index variable because e had to. If you multiply a sum of, for instance, seven terms by itself, the result ill be a sum ith 49 terms you ve got to take all possible products of any of the terms ith any of the other terms. So far, e are assuming that e are orking in a real inner product space.use (1.5) and (1.6) to simplify (1.): v j α j e j = v, v j v, α j e j j α j e j, v + j,k α j e j, α k e k = v, v j α j v, e j + j,k α j α k e j, e k = v, v j α j v, e j + j α j by 1.1) 88

8 = j ( α j α j v, e j ) + v We complete the square, using formula (11.1): = j ( α j α j v, e j + v, e j v, e j ) + v = j (α j v, e j ) j v, e j + v (1.3) We re trying to minimize this quantity, and e get to choose the α j any ay e ant to. It is clear that the first summation on the right hand side of (1.3), being the sum of squares, is alays greater than or equal to zero but e can make it zero if e just choose the α j s right. The right choice is to let α j be equal to v, e j. These values for the coefficients are called (for reasons hich ill eventually become clear) the generalized Fourier coefficients for v ith respect to this orthonormal set. Since the left hand side of (1.) is the square of a norm, it must alays be greater than or equal to zero. If e let α j be equal to v, e j for each j, the remaining portion of the right hand side of (1.3) must be nonnegative. That is, e have the folloing inequality, generally knon as Bessel s inequality: v, e j v (1.4) j If e are able to rite v as a linear combination of the ej s, then the minimum value of the left hand side of (1.) ould be zero and the inequality in (1.4) ould actually be an equality. What if the orthonormal set is actually an infinite set rather than a finite set? Then such sums as appear in (1.) and (1.4) ould have to be interpreted as infinite series. Fortunately, the very orkings of this problem notably Bessel s inequality help to assure us that these series converge in some appropriate sense. (Actually, to get everything that e might ask for in terms of these series being meaningful, e ll have to have our inner product space be a complete metric space, hich makes it a Hilbert space unfortunately, this is not the same meaning of the ord complete as e are about to use belo.) Bessel s inequality ill alays be true, and the norm of the difference beteen v and our linear combination of the ej s ill alays be minimized if e choose our coefficients to be the generalized Fourier coefficients. A ne issue no arises: does our orthonormal set have enough elements in it that e can rite any vector v as the limit of linear combinations of that is, as an infinite series based on that set? If so, e call the orthonormal set complete. (This is also knon as having an orthonormal basis.) It happens and the calculations above provide the frameork for this argument, too that an orthonormal set is complete if and only if the only vector orthogonal to every element in it is the zero vector. Let s summarize our findings: Fourier coefficients: v j Bessel s inequality: α j e j is minimized if each α j = v, e j (1.5) v, e j v (1.6) j 89

9 If the orthonormal set is also complete, then: Generalized Fourier series: v = k v, e j e j (1.7) Generalized arseval s identity: v, e j = v (1.8) j 1.7. Orthonormal sets in a complex inner product space No e suppose that {e j } is a (finite, for the time being) orthonormal set in a complex inner product space. Without further ado, let s repeat hat as in the last section: v j α j e j = v j α j e j, v k α k e j = v, v j v, α j e j j α j e j, v j,k α j e j, α k e k = v, v j α j v, e j j α j v, e j + j,k α j α k e j, e k = v, v j α j v, e j j α j v, e j + j α j = j ( ) α j α j v, e j α j v, e j + v We complete the square, using formula (11.7): = ( α j α j v, e j α j v, e j + v, e j v, e j ) + v j = j α j v, e j j v, e j + v (1.9) No e are ready to dra conclusions from this, exactly as before. Formulas (1.5), (1.6), (1.7), and (1.8) are also valid ithout change in a complex inner product space. Of course, e ere clever enough to include some ell-chosen absolute values in these statements! 1.8. Square integrable functions on T Consider a function f (real or complex valued) that is defined on the line so as to be periodic of period. We call such a function square integrable, and say that it belongs to L (T ),provided the folloing integral converges: 1 f(x) dx < 0 90

10 Just to give you a taste of this condition it doesn t require that the function be bounded, but it is harder to satisfy than mere integrability, or even absolute integrability. As an example, take the function f(x) = 1 for 0 < x [ x. This function is integrable on, ] (and hence absolutely integrable since it is positive), but if e square it, e get 1, hich is not integrable. x We say that this function belongs to L 1 but not to L. We no define the inner product. lease recognize that our decision to divide out front by represents one of several possible notational choices, and may not necessarily be reflected in other orks. The inner product for real-valued functions: If f, g L (T ), then f, g = 1 The inner product for complex-valued functions: If f, g L (T ), then f, g = 1 In either case, the norm is as follos: f(x)g(x) dx (1.30) f(x)g(x) dx (1.31) f = f, f = ( 1 ) 1 f(x) dx (1.3) The Cauchy-Scharz inequality in this case reads as follos: 1 ( f(x)g(x) dx 1 A quick corollary of this is that ) 1 ( f(x) 1 dx g(x) dx) 1 = f g (1.33) f 1 = 1 f(x) dx ( 1 ) 1 ( f(x) 1 dx 1 dx) 1 f (1.34) 1.9. Fourier series and orthogonal expansions { 1, cos ( πx ), sin ( πx ), cos ( πx ), sin ( πx ), cos ( 3 πx ), sin ( 3 πx ) },... is an orthogonal set in the real inner product space L (T ), and almost but not quite an orthonormal set. The specific problem ith being orthonormal is the normalization: the function 1 has norm equal to 1, but all of the other functions in this set have norm equal to 1. We could chase through all of the consequences of that factor, but rather than give you the details, e ll just give the results. If e ant to approximate a square integrable function f by an N th degree trigonometric polynomial, the the closest e can come in the L norm the least squares or least root mean square approximation is to let the coefficients of this polynomial be exactly the Fourier coefficients. That 91

11 is, e let this trigonometric polynomial be the Nth partial sum of the Fourier series for f. One consequence of this is by (1.6) Bessel s inequality: hich can also be ritten as a[0] + a[0] N ( a[k] + b[k] ) f (1.35) N ( a[k] + b[k] ) f(x) dx (1.36) But then, it turns out that this orthogonal set is complete. We on t prove that, but basically, our previous convergence theorems for Fourier series make this inevitable. That being the case, e can say that the Fourier series of any square integrable function alays converges to that function in the L sense. Furthermore, by (1.8), e have arseval s identity: a[0] ( a[k] + b[k] ) = f (1.37) a[0] ( + a[k] + b[k] ) = f(x) dx (1.38) Let s try to repeat this for the complex case. This time, consider the set {e πikx/ } k= of complex valued functions on T. This turns out to be an orthonormal set in fact, demonstrating that fact is far easier for this case than for the real case. This is also a complete orthonormal set e couldn t possibly have a different result for the complex case than for the real case so e have both a Bessel s inequality and a arseval s identity for this case, too. Bessel s inequality: arseval s identity: N f[k] f (1.39) k= N f[k] = f (1.40) k= The trigonometric orthogonal set on N On the polygon N, the set of functions e k [n] = e πikn/n form an orthogonal set ith respect to the usual complex inner product. That is, { N 1 N 1 e j, e k = e j [n]e k [n] = e πijn/n e πikn/n N if j = k = (1.41) 0 otherise n=0 n=0 Equation (1.41) is just a restatement of equations (4.1) and (4.6). The factor of N that appears in it means that the vectors are orthogonal but not quite orthonormal. Let s put this another ay. Suppose e have an N-dimensional complex inner product space ith the standard inner product ith respect to that basis. (The set of all complex-valued functions on N is just such a space.) We 9

12 can then rite any vector as an N 1 column matrix. Suppose e have a set of M such vectors. Create an N M matrix A hose M columns are these M vectors. If M = N e ill have a square matrix. Ho could e tell if these M vectors ere independent? This ould be a question about the rank of the matrix A. If its rank is M, e have an independent set. In the M = N square matrix case, the N vectors are independent if and only if the determinant of A is not zero. No ho can e tell if the set of vectors is orthogonal? To do this, let A be the complex conjugate of the transpose of A, and compute the matrix product A A. This ill be the M M matrix hose (j, k)th entry is precisely the inner product of the jth and kth vectors of our set. The statement that the set is orthogonal is precisely the statement that A A is diagonal: that all of its entries are zero except those on the main diagonal hich are not zero. The statement that our set is orthonormal is precisely the statement that A A = I, the identity matrix. If A is a square matrix such that A A = I, e call A a unitary matrix. Use this language to express (1.41). Let F be the N N matrix hose columns are the vectors e j named above. In other ords, the (j, k)th entry of F is e πijk/n = ζ jk. Then F is almost but not quite a unitary matrix: F F = NI. So e have an orthogonal but not quite orthonormal set. What can e say about this in general? Suppose that {e j } is a finite orthogonal (but not necessarily orthonormal) set in a complex inner product space, and that v is an arbitrary vector in that space. We ish to find the coefficients α j such that the norm v α j e j is minimized. This minimum is achieved if and only if j α j = v, e j e j, e j (1.4) Bessel s inequality in this case turns out to be j v, e j e j, e j v (1.43) ith equality (arseval s identity) in the case of the orthogonal set being complete. We may easily apply (1.4) and (1.43) to the case of the trigonometric basis for functions in N. The coefficients in (1.4) turn out to be: f, e j e j, e j = 1 N 1 f[n]e πijn/n = N f[j] (1.44) n=0 The set of all N of these trigonometric functions is a set of N orthogonal, hence independent, vectors. Therefore, they span the N-dimensional vector space of all functions on the polygon and are a complete orthogonal set. Equality necessarily holds in (1.43). A careful orking out of the consequences of (1.43) leads us to arseval s identity for this instance: N f = N N 1 k=0 f[k] = N 1 n=0 f[n] = f (1.45) Exercises 1.1. By imitating the derivations around equation (1.9), prove (1.4) and (1.43). 93

13 1.. Which of the folloing Fourier series represent square-integrable functions on T π, also knon as [ π, π]? What else can you say about the functions they represent? (a) (b) (c) (d) k= sin kx k = sin kx k = cos kx k = k= k 0 k= k 0 k= k 0 e ikx ik k e ikx e ikx k ik r k e ikx, here 0 r < 1 Supplemental exercises: 1.3. Use the Fourier series of pieceise-polynomial functions on T and either arseval s identity 1 or the synthesis equation at ell-chosen points to calculate k, 1 k 4, and 1 k 6. One possibility: Start ith f 1 (x) = 1 x on the interval (0, 1), extended to be periodic of period 1. Build a sequence of functions f n such that f n+1(x) = f n (x). (That is, f n+1 is an antiderivative of f n.) Choose the constant of integration so that f n+1 [0] = 1 0 f n+1 (x) dx = Find a ay to automate the calculations in exercise 1.3, using a computer to help. Among other things, you ll need the ability to ork ith arbitrary precision rational numbers (hence arbitrarily long integers). DERIVE, MALE, and MATHEMATICA have this capability. 1 Calculate the exact value of. (Why 6? Because that as as far as Euler got, orking k6 the problem by hand.) 94