Since both of these one-sided limits are equal, they must also both equal zero. To prove the Extreme Value Theorem, suppose a continuous function f does not achieve a maximum value on a compact set. 809.15 935.92 981.04 702.19 647.82 717.8 719.93 1135.11 818.86 764.37 823.14 769.85 That is to say, $f$ attains its maximum on $[a,b]$. Theorem 7.3 (Mean Value Theorem MVT). when x > K we have that f (x) > M. Unformatted text preview: Intermediate Value Property for Functions with Antiderivatives Theorem 5.3. let Suppose that ϕ is di erentiable at each point of the interval I and ϕ (x) = f (x) for all x ∈ I. /LastChar 255 Letfi =supA. 27 0 obj If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /StemV 80 /CapHeight 683.33 /Subtype /Type1 Then the image D as defined in the lemma above is compact. In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. 0 675.93 937.5 875 787.04 750 879.63 812.5 875 812.5 875 812.5 656.25 625 625 937.5 21 0 obj /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /ff /fi /fl /ffi /ffl /dotlessi The proof of the extreme value theorem is beyond the scope of this text. /FontBBox [-116 -350 1278 850] endobj Prove using the definitions that f achieves a minimum value. >> About the Author James Lowman is an applied mathematician currently working on a Ph.D. in the field of computational fluid dynamics at the University of Waterloo. /Encoding 7 0 R /Filter [/FlateDecode] 819.39 934.07 838.69 724.51 889.43 935.62 506.3 632.04 959.93 783.74 1089.39 904.87 /FirstChar 33 result for constrained problems. If f(x) has an extremum on an open interval (a,b), then the extremum occurs at a critical point. /XHeight 444.4 k – ε < f (c) < k + ε. /FontDescriptor 9 0 R This is the Weierstrass Extreme Value Theorem. Now we turn to Fact 1. /Name /F4 /FontName /TFBPDM+CMSY7 /Widths [1138.89 585.32 585.32 1138.89 1138.89 1138.89 892.86 1138.89 1138.89 708.34 465.63 489.59 476.97 576.16 344.51 411.81 520.6 298.38 878.01 600.23 484.72 503.13 >> 769.85 769.85 769.85 769.85 708.34 708.34 523.81 523.81 523.81 523.81 585.32 585.32 stream 630.96 323.41 354.17 600.2 323.41 938.5 630.96 569.45 630.96 600.2 446.43 452.58 Proof of Fermat’s Theorem. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Proof: Let f be continuous, and let C be the compact set on which we seek its maximum and minimum. endobj /XHeight 444.4 /Type /FontDescriptor For every ε > 0. << 889.45 616.08 818.35 688.52 978.64 646.5 782.15 871.68 791.71 1342.69 935.57 905.84 446.41 451.16 468.75 361.11 572.46 484.72 715.92 571.53 490.28 465.05 322.46 384.03 /Ascent 750 333.33 277.78 500 500 500 500 500 500 500 500 500 500 500 277.78 277.78 277.78 777.78 0 0 0 0 0 0 575] By the Extreme Value Theorem there must exist a value in that is a maximum. /Descent -250 772.4 639.7 565.63 517.73 444.44 405.9 437.5 496.53 469.44 353.94 576.16 583.34 602.55 Because f(x) is continuous on [a, b], by the Extreme Value Theorem, we know that f(x) will have a minimum somewhere on [a, b]. The Extreme Value Theorem, sometimes abbreviated EVT, says that a continuous function has a largest and smallest value on a closed interval.This is used to show thing like: There is a way to set the price of an item so as to maximize profits. endobj 24 0 obj (Extreme Value Theorem) If $f$ is a continuous function on a closed bounded interval $ [a,b],$ then $f$ must attain an absolute maximum value $f (s)$ and an absolute minimum value $f (t)$ at some numbers $s$ and $t$ in $ [a,b].$ /Encoding 7 0 R /Type /Font >> We look at the proof for the upper bound and the maximum of f. 750 611.11 277.78 500 277.78 500 277.78 277.78 500 555.56 444.45 555.56 444.45 305.56 Define a new function $g$ by $\displaystyle{g(x) = \frac{1}{M-f(x)}}$. 0 0 0 339.29] We show that, when the buyer’s values are independently distributed /XHeight 430.6 575 638.89 606.94 473.61 453.61 447.22 638.89 606.94 830.55 606.94 606.94 511.11 /Ascent 750 708.34 1138.89 1138.89 1138.89 892.86 329.37 1138.89 769.85 769.85 1015.88 1015.88 /Subtype /Type1 0 693.75 954.37 868.93 797.62 844.5 935.62 886.31 677.58 769.84 716.89 880.04 742.68 /Subtype /Type1 /FontFile 8 0 R That leaves as the only possibility that there is some $c$ in $[a,b]$ where $f(c) = M$. /Widths [350 602.78 958.33 575 958.33 894.44 319.44 447.22 447.22 575 894.44 319.44 585.32 585.32 585.32 339.29 339.29 892.86 585.32 892.86 585.32 610.07 859.13 863.18 22 0 obj /Ascent 750 The extreme value theorem itself was first proved by the Bohemian mathematician and philosopher Bernard Bolzano in 1830, but his book, Function Theory, was only published a hundred years later in 1930. Extreme Value Theorem: If a function is continuous in a closed interval , with the maximum of at and the minimum of at then and are critical values of Proof: The proof follows from Fermat’s theorem and is left as an exercise for the student. The result was also discovered later by Weierstrass in 1860. /LastChar 255 7 0 obj 575 1149.99 575 575 0 691.66 958.33 894.44 805.55 766.66 900 830.55 894.44 830.55 Since $f$ is continuous on $[a,b]$, we know it must be bounded on $[a,b]$ by the Boundedness Theorem. 1188.88 869.44 869.44 702.77 319.44 602.78 319.44 575 319.44 319.44 559.02 638.89 /Widths [719.68 539.73 689.85 949.96 592.71 439.24 751.39 1138.89 1138.89 1138.89 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 864.58 849.54 1162.04 849.54 849.54 687.5 312.5 581.02 312.5 562.5 312.5 312.5 546.88 One has to use the fact that is a real closed field, but since there are lots of real closed fields, one usually defines in a fundamentally analytic way and then proves the intermediate value theorem, which shows that is a real closed field. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1018.52 1143.52 875 312.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded interval. /Flags 68 /Type /Encoding This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already. /FirstChar 33 /ItalicAngle 0 First we will show that there must be a finite maximum value for f (this << 569.45 569.45 323.41 323.41 323.41 876.99 538.69 538.69 876.99 843.26 798.62 815.48 /FontName /PJRARN+CMMI10 Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. The Extreme Value Theorem tells us that we can in fact find an extreme value provided that a function is continuous. endobj 28 0 obj Note that $g(x) \gt 0$ for every $x$ in $[a,b]$ and $g$ is continuous on $[a,b]$, and thus also bounded on this interval (again by the Boundedness theorem). /FontDescriptor 12 0 R 462.3 462.3 462.3 1138.89 1138.89 478.18 619.66 502.38 510.54 594.7 542.02 557.05 539.19 431.55 675.44 571.43 826.44 647.82 579.37 545.81 398.65 441.97 730.11 585.32 472.22 472.22 777.78 750 708.34 722.22 763.89 680.56 652.78 784.72 750 361.11 513.89 >> >> /BaseFont /PJRARN+CMMI10 If a function $f$ is continuous on $[a,b]$, then it attains its maximum and minimum values on $[a,b]$. /FontDescriptor 24 0 R 342.59 875 531.25 531.25 875 849.54 799.77 812.5 862.27 738.43 707.18 884.26 879.63 butions requires the proof of novel extreme value theorems for such distributions. 694.45 666.67 750 722.22 777.78 722.22 777.78 722.22 583.34 555.56 555.56 833.34 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 769.85 769.85 892.86 892.86 523.81 523.81 523.81 708.34 892.86 892.86 892.86 0 892.86 /Flags 4 /Ascent 750 (a) Find the absolute maximum and minimum values of f (x) 4x2 12x 10 on [1, 3]. Indeed, complex analysis is the natural arena for such a theorem to be proven. Proof of the Extreme Value Theorem Theorem: If f is a continuous function defined on a closed interval [a;b], then the function attains its maximum value at some point c contained in the interval. Proof: There will be two parts to this proof. This theorem is sometimes also called the Weierstrass extreme value theorem. endobj 625 500 625 513.31 343.75 562.5 625 312.5 343.75 593.75 312.5 937.5 625 562.5 625 The standard proof of the first proceeds by noting that f is the continuous image of a compact set on the interval [a,b], so it must itself be compact. Proof of the Extreme Value Theorem. /Type /FontDescriptor Then $f(x) \lt M$ for all $x$ in $[a,b]$. /acute /caron /breve /macron /ring /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /Ascent 750 We need Rolle’s Theorem to prove the Mean Value Theorem. /Name /F5 /CapHeight 683.33 /Subtype /Type1 << Since the function is bounded, there is a least upper bound, say M, for the range of the function. /CapHeight 686.11 >> Hence by the Intermediate Value Theorem it achieves a … /Descent -951.43 /Widths [622.45 466.32 591.44 828.13 517.02 362.85 654.17 1000 1000 1000 1000 277.78 3 It states the following: If a function f (x) is continuous on a closed interval [ a, b ], then f (x) has both a maximum and minimum value on [ a, b ]. 418.98 581.02 880.79 675.93 1067.13 879.63 844.91 768.52 844.91 839.12 625 782.41 Sketch of Proof. /StemV 80 869.44 511.11 597.22 830.55 894.44 575 1041.66 1169.44 894.44 319.44 0 0 0 0 0 0 Both proofs involved what is known today as the Bolzano–Weierstrass theorem. 680.56 970.14 803.47 762.78 642.01 790.56 759.29 613.2 584.38 682.78 583.33 944.45 446.43 630.96 600.2 815.48 600.2 600.2 507.94 569.45 1138.89 569.45 569.45 0 706.35 /FirstChar 33 /CapHeight 683.33 /FontName /YNIUZO+CMR7 /Name /F3 /FontFile 20 0 R >> 500 530.9 750 758.51 714.72 827.92 738.2 643.06 786.25 831.25 439.58 554.51 849.31 It is a special case of the extremely important Extreme Value Theorem (EVT). /Encoding 7 0 R It tends to zero in the limit, so we exploit that in this proof to show the Fundamental Theorem of Calculus Part 2 is true. /FirstChar 33 Given that $g$ is bounded on $[a,b]$, there must exist some $K \gt 0$ such that $g(x) \le K$ for every $x$ in $[a,b]$. endobj /Type /FontDescriptor endobj The proof that $f$ attains its minimum on the same interval is argued similarly. Weclaim that thereisd2[a;b]withf(d)=fi. /BaseFont /UPFELJ+CMBX10 /ItalicAngle 0 /Descent -250 /Descent -250 State where those values occur. endobj So since f is continuous by defintion it has has a minima and maxima on a closed interval. 500 555.56 277.78 305.56 527.78 277.78 833.34 555.56 500 555.56 527.78 391.67 394.45 f (c) < (f (x) + ε) ≤ (k + ε) ——– (1) Similarly, values of x between c and c + δ that are not contained in A, such that. /ItalicAngle 0 /FontDescriptor 18 0 R /LastChar 255 383.33 319.44 575 575 575 575 575 575 575 575 575 575 575 319.44 319.44 350 894.44 647.77 600.08 519.25 476.14 519.84 588.6 544.15 422.79 668.82 677.58 694.62 572.76 493.98 437.5 570.03 517.02 571.41 437.15 540.28 595.83 625.69 651.39 0 0 0 0 0 0 636.46 500 0 615.28 833.34 762.78 694.45 742.36 831.25 779.86 583.33 666.67 612.22 /FontBBox [-119 -350 1308 850] /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /FontName /IXTMEL+CMMI7 First, it follows from the Extreme Value Theorem that f has an absolute maximum or minimum at a point c in (a, b). /FontName /NRFPYP+CMBX12 We now build a basic existence result for unconstrained problems based on this theorem. (a) Find the absolute maximum and minimum values of x g(x) x2 2000 on (0, +∞), if they exist. /XHeight 430.6 /Widths [323.41 569.45 938.5 569.45 938.5 876.99 323.41 446.43 446.43 569.45 876.99 It may then be shown that: f 0 (c) = lim h → 0 f (c + h)-f (c) h = 0, using that fact that if f (c) is an absolute extremum, then f (c + h)-f (c) h is both ≤ 0 and ≥ 0. << Suppose there is no such $c$. /XHeight 430.6 /LastChar 255 /FontFile 14 0 R 0 0 646.83 646.83 769.85 585.32 831.35 831.35 892.86 892.86 708.34 917.6 753.44 620.18 828.47 580.56 682.64 388.89 388.89 388.89 1000 1000 416.67 528.59 429.17 432.76 520.49 /Type /Font 868.93 727.33 899.68 860.61 701.49 674.75 778.22 674.61 1074.41 936.86 671.53 778.38 833.34 277.78 305.56 500 500 500 500 500 750 444.45 500 722.22 777.78 500 902.78 /CapHeight 683.33 Therefore proving Fermat’s Theorem for Stationary Points. Proof of the Intermediate Value Theorem; The Bolzano-Weierstrass Theorem; The Supremum and the Extreme Value Theorem; Additional Problems; 11 Back to Power Series. /Subtype /Type1 << << /XHeight 430.6 1013.89 777.78 277.78 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /FontFile 17 0 R /FirstChar 33 /FontDescriptor 15 0 R /FontFile 11 0 R Theorem 6 (Extreme Value Theorem) Suppose a < b. We prove the case that $f$ attains its maximum value on $[a,b]$. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 << /Flags 68 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Name /F6 We needed the Extreme Value Theorem to prove Rolle’s Theorem. /Widths [277.78 500 833.34 500 833.34 777.78 277.78 388.89 388.89 500 777.78 277.78 /Flags 68 /BaseFont /YNIUZO+CMR7 557.33 668.82 404.19 472.72 607.31 361.28 1013.73 706.19 563.89 588.91 523.6 530.43 /BaseFont /NRFPYP+CMBX12 It is necessary to find a point d in [ a , b ] such that M = f ( d ). >> A continuous function (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue). Thus, before we set off to find an absolute extremum on some interval, make sure that the function is continuous on that interval, otherwise we may be hunting for something that does not exist. 323.41 384.92 323.41 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 569.45 /ItalicAngle -14 388.89 555.56 527.78 722.22 527.78 527.78 444.45 500 1000 500 500 0 625 833.34 777.78 >> If there is some $c$ in $[a,b]$ where $f(c) = M$ there is nothing more to show -- $f$ attains its maximum on $[a,b]$. Theorem \(\PageIndex{2}\): Extreme Value Theorem (EVT) Suppose \(f\) is continuous on \([a,b]\). /oslash /AE /OE /Oslash 161 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /StemV 80 Thus for all in . The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at some point in that interval. 19 0 obj (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set. endobj endobj /CapHeight 683.33 /Type /FontDescriptor Examples 7.4 – The Extreme Value Theorem and Optimization 1. Hence, the theorem is proved. /BaseFont /JYXDXH+CMR10 /Type /FontDescriptor This makes sense because the function must go up (as) and come back down to where it started (as). 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /FontBBox [-100 -350 1100 850] which implies (upon multiplication of both sides by the positive $M-f(x)$, followed by a division on both sides by $K$) that for every $x$ in $[a,b]$: This is impossible, however, as it contradicts the assumption that $M$ was the least upper bound! 860.12 767.86 737.11 883.93 843.26 412.7 583.34 874.01 706.35 1027.78 843.26 876.99 /ItalicAngle 0 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /dieresis] The extreme value theorem is used to prove Rolle's theorem. 0 0 0 0 0 0 277.78] /StemV 80 %PDF-1.3 /suppress /dieresis /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /Name /F2 >> << /Descent -250 As a byproduct, our techniques establish structural properties of approximately-optimal and near-optimal solutions. /Name /F7 /Length 3528 /Descent -250 << >> Suppose the least upper bound for $f$ is $M$. 462.3 462.3 339.29 585.32 585.32 708.34 585.32 339.29 938.5 859.13 954.37 493.56 The extreme value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930. endobj /FontBBox [-134 -1122 1477 920] 13 0 obj So there must be a maximum somewhere. 938.5 876.99 781.75 753.97 843.26 815.48 876.99 815.48 876.99 815.48 677.58 646.83 25 0 obj /Flags 4 The extreme value theorem: Any continuous function on a compact set achieves a maximum and minimum value, and does so at specific points in the set. 0 892.86] /StemV 80 543.05 543.05 894.44 869.44 818.05 830.55 881.94 755.55 723.61 904.16 900 436.11 /Type /Font Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. Proof of the extreme value theorem By the boundedness theorem, f is bounded from above, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) M of f exists. << /BaseEncoding /WinAnsiEncoding https://www.khanacademy.org/.../ab-5-2/v/extreme-value-theorem 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 << /Type /FontDescriptor >> /FirstChar 33 16 0 obj << ThenA 6= ;and, by theBounding Theorem, A isboundedabove andbelow. /Differences [0 /Gamma /Delta /Theta /Lambda /Xi /Pi /Sigma /Upsilon /Phi /Psi /Omega /LastChar 255 We will first show that \(f\) attains its maximum. /FirstChar 33 /Flags 4 18 0 obj 1138.89 585.32 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /StemV 80 /FontName /UPFELJ+CMBX10 The Mean Value Theorem for Integrals. /ItalicAngle -14 /FontDescriptor 27 0 R Then there exists \(c\), \(d ∈ [a,b]\) such that \(f(d) ≤ f(x) ≤ f(c)\), for all \(x ∈ [a,b]\). 1138.89 339.29 339.29 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 585.32 /Type /Font Sketch of Proof. /Descent -250 /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef /.notdef 339.29 892.86 585.32 892.86 585.32 892.86 892.86 892.86 892.86 892.86 892.86 892.86 /Encoding 7 0 R The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions. 519.84 668.05 592.71 661.99 526.84 632.94 686.91 713.79 755.96 0 0 0 0 0 0 0 0 0 /Subtype /Type1 12 0 obj 894.44 830.55 670.83 638.89 638.89 958.33 958.33 319.44 351.39 575 575 575 575 575 /FontDescriptor 21 0 R 39 /quoteright 60 /exclamdown 62 /questiondown 92 /quotedblleft 94 /circumflex /dotaccent f (c) > (f (x) – ε) > (k − ε) ——– (2) Combining both the inequality relations, obtain. /BaseFont /IXTMEL+CMMI7 Among all ellipses enclosing a fixed area there is one with a … /Subtype /Type1 /Type /Font /FontBBox [-103 -350 1131 850] endobj If f : [a;b] !R, then there are c;d 2[a;b] such that f(c) •f(x) •f(d) for all x2[a;b]. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Proof LetA =ff(x):a •x •bg. /cedilla /germandbls /ae /oe /oslash /AE /OE /Oslash /suppress 34 /quotedblright endobj 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 562.5] 569.45] Suppose that is defined on the open interval and that has an absolute max at . 767.86 876.99 829.37 630.96 815.48 843.26 843.26 1150.8 843.26 843.26 692.46 323.41 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 << /quoteleft 123 /endash /emdash /hungarumlaut /tilde /dieresis /Gamma /Delta /Theta 937.5 312.5 343.75 562.5 562.5 562.5 562.5 562.5 849.54 500 574.07 812.5 875 562.5 /XHeight 430.6 /FontFile 26 0 R << Depending on the setting, it might be needed to decide the existence of, and if they exist then compute, the largest and smallest (extreme) values of a given function. 10 0 obj xڵZI����WT|��%R��$@��������郦J�-���)�f��|�F�Zj�s��&������VI�$����m���7ߧ�4��Y�?����I���ԭ���s��Css�Өy������sŅ�>v�j'��:*�G�f��s�@����?V���RjUąŕ����g���|�C��^����Cq̛G����"N��l$��ӯ]�9��no�ɢ�����F�QW�߱9R����uWC'��ToU��� W���sl��w��3I�뛻���ݔ�T�E���p��!�|�dLn�ue���֝v��zG�䃸� ���)�+�tlZ�S�Q���Q7ݕs�s���~�����s,=�3>�C&�m:a�W�h��*6�s�K��C��r��S�;���"��F/�A��F��kiy��q�c|s��"��>��,p�g��b�s�+P{�\v~Ξ2>7��u�SW�1h����Y�' _�O���azx\1w��%K��}�[&F�,pЈ�h�%"bU�o�n��M���D���mٶoo^�� *`��-V�+�A������v�jv��8�Wka&�Q. /FontName /JYXDXH+CMR10 /Type /Font /Name /F1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Another mathematician, Weierstrass, also discovered a proof of the theorem in 1860. >> /Phi /Psi /.notdef /.notdef /Omega /ff /fi /fl /ffi /ffl /dotlessi /dotlessj /grave 569.45 323.41 569.45 323.41 323.41 569.45 630.96 507.94 630.96 507.94 354.17 569.45

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