5Mathematics I Study Guide: Series, Limits, and Binomial Theorem

5. Übung zur Vorlesung Mathematik I (SoSe 26) - 25.5.26

Aufgabe 1: Babylonisches Wurzelziehen (Babylonian Root Extraction)

  • Context and Objective: In the lecture, an iterative procedure was introduced to calculate the square root of 3. This exercise requires applying a modified version of this method to calculate the cubic root of 3 (33\sqrt[3]{3}).
  • Concept Development:
    • Instead of considering a square, a cube (cube) must be considered.
    • The goal is to find the side length of a cube with a volume of 3.
    • If ana_n is the current approximation for the side length, the corresponding "depth" or secondary dimension to maintain the volume is bn=3an2b_n = \frac{3}{a_n^2}.
  • Iteration Formula:
    • The next value in the sequence, an+1a_{n+1}, is obtained by taking the average of the current side length and the secondary dimension:
    • an+1=an+bn2=an+3an22a_{n+1} = \frac{a_n + b_n}{2} = \frac{a_n + \frac{3}{a_n^2}}{2}
  • Iteration Calculations:
    • Starting Point: Let a0=1a_0 = 1.
    • Step 1:
      • a1=1+312=42=2a_1 = \frac{1 + \frac{3}{1}}{2} = \frac{4}{2} = 2
    • Step 2:
      • a2=2+3222=2+342=1142=118=1.375a_2 = \frac{2 + \frac{3}{2^2}}{2} = \frac{2 + \frac{3}{4}}{2} = \frac{\frac{11}{4}}{2} = \frac{11}{8} = 1.375
    • Step 3:
      • a3=118+3(11/8)22a_3 = \frac{\frac{11}{8} + \frac{3}{(11/8)^2}}{2}
      • The term for b2b_2 is 312164=192121\frac{3}{\frac{121}{64}} = \frac{192}{121}.
      • a3=118+1921212=1331+15361936=28671936×12a_3 = \frac{\frac{11}{8} + \frac{192}{121}}{2} = \frac{1331 + 1536}{1936} = \frac{2867}{1936} \times \frac{1}{2}. (Correction based on transcript arithmetic: a3=28671936×12a_3 = \frac{2867}{1936} \times \frac{1}{2} seems to be simplified in the text as approximately 1.4811.481).
  • Accuracy Check: The value 1.4811.481 deviates by approximately 3%3 \% from the exact solution of 33≈1.442\sqrt[3]{3} ≈ 1.442.

Aufgabe 2: Reihen (Series)

  • Objective: Calculate the result of the series S(n)=∑k=1nkS(n) = \frac{\sum}{k=1}^n k (where n∈N,n>0n ∈ ℕ, n > 0) using an index transformation.
  • Methodology:
    • Replace kk with the transformation k=n+1−jk = n + 1 - j.
    • This transformation merely changes the order of summation and does not change the total sum.
  • Proof Steps:
    • Let S(n)=∑k=1nkS(n) = \frac{\sum}{k=1}^n k.
    • Applying the transformation: S(n)=∑j=1n(n+1−j)S(n) = \frac{\sum}{j=1}^n (n + 1 - j).
    • Using the linearity of summation, the sum over jj can be split:
      • S(n)=∑j=1n(n+1)−∑j=1njS(n) = \frac{\sum}{j=1}^n (n + 1) - \frac{\sum}{j=1}^n j
    • Since (n+1)(n+1) is constant relative to jj, the first sum is n×(n+1)n \times (n + 1).
    • The second sum, ∑j=1nj\frac{\sum}{j=1}^n j, is identical to the original sum S(n)S(n).
    • Therefore: S(n)=n×(n+1)−S(n)S(n) = n \times (n + 1) - S(n).
  • Final Formula Derivation:
    • 2×S(n)=n(n+1)2 \times S(n) = n(n + 1)
    • S(n)=∑k=1nk=n(n+1)2S(n) = \frac{\sum}{k=1}^n k = \frac{n(n + 1)}{2}

Aufgabe 3: Grenzwertbildung (Limits)

  • Problem a: Calculate lim⁡n→∞1n2+4×cos(1n+1)\lim_{n \to \infty} \frac{1}{n^2} + 4 \times \text{cos}(\frac{1}{n+1}).
    • Calculation: As nn approaches infinity, 1n2\frac{1}{n^2} goes to 00 and 1n+1\frac{1}{n+1} goes to 00.
    • 0+4×cos(0)=0+4×1=40 + 4 \times \text{cos}(0) = 0 + 4 \times 1 = 4.
  • Problem b: Calculate lim⁡n→∞cos(n)en\lim_{n \to \infty} \frac{\text{cos}(n)}{e^n}.
    • Tipp: Represent the term as a product: cos(n)×e−n\text{cos}(n) \times e^{-n}.
    • Argument: The function cos(n)\text{cos}(n) is bounded (its values stays between −1-1 and 11).
    • The exponential function part satisfies lim⁡n→∞e−n=0\lim_{n \to \infty} e^{-n} = 0.
    • A bounded sequence multiplied by a null sequence results in a null sequence. Thus, the limit is 00.
  • Problem c: Calculate lim⁡n→∞2n2+n+1n2+n+2\lim_{n \to \infty} \frac{2n^2 + n + 1}{n^2 + n + 2}.
    • Tipp: Factor out n2n^2 from the numerator and denominator.
    • Calculation:
      • n2×(2+1n+1n2)n2×(1+1n+2n2)=2+1n+1n21+1n+2n2\frac{n^2 \times (2 + \frac{1}{n} + \frac{1}{n^2})}{n^2 \times (1 + \frac{1}{n} + \frac{2}{n^2})} = \frac{2 + \frac{1}{n} + \frac{1}{n^2}}{1 + \frac{1}{n} + \frac{2}{n^2}}
    • As n→∞n \to \infty, all terms with nn in the denominator approach 00.
    • Result: 2+0+01+0+0=2\frac{2 + 0 + 0}{1 + 0 + 0} = 2.

Aufgabe 4: Binomialkoeffizienten (Binomial Coefficients)

  • Identity a: Prove (nn−k)=(nk)\binom{n}{n-k} = \binom{n}{k}.
    • Proof: By definition, (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}.
    • Using commutativity of multiplication: n!k!(n−k)!=n!(n−k)!k!\frac{n!}{k!(n-k)!} = \frac{n!}{(n-k)!k!}.
    • This fulfills the formula for (nn−k)\binom{n}{n-k}.
  • Identity b: Prove (nk−1)+(nk)=(n+1k)\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}.
    • Proof: Apply the definition of binomial coefficients to both terms:
      • (nk−1)+(nk)=n!(k−1)!(n−(k−1))!+n!k!(n−k)!=n!(k−1)!(n−k+1)!+n!k!(n−k)!\binom{n}{k-1} + \binom{n}{k} = \frac{n!}{(k-1)!(n-(k-1))!} + \frac{n!}{k!(n-k)!} = \frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{k!(n-k)!}
    • To add these, bring them to a common denominator by multiplying the first term by kk\frac{k}{k} and the second term by n−k+1n−k+1\frac{n-k+1}{n-k+1}:
      • k×n!k!(n−k+1)!+(n−k+1)×n!k!(n−k+1)!\frac{k \times n!}{k!(n-k+1)!} + \frac{(n-k+1) \times n!}{k!(n-k+1)!}
    • Combine numerators:
      • (k+n−k+1)×n!k!(n−k+1)!=(n+1)×n!k!(n−k+1)!=(n+1)!k!((n+1)−k)!\frac{(k + n - k + 1) \times n!}{k!(n-k+1)!} = \frac{(n+1) \times n!}{k!(n-k+1)!} = \frac{(n+1)!}{k!((n+1)-k)!}
    • The resulting expression is the definition of (n+1k)\binom{n+1}{k}.

Aufgabe 5: Binomialtheorem (Binomial Theorem)

  • Formula: (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \frac{\sum}{k=0}^n \binom{n}{k} a^{n-k} b^k
  • Problem a: Calculate 1023102^3.
    • (100+2)3=1003+3×1002×2+3×100×22+23(100 + 2)^3 = 100^3 + 3 \times 100^2 \times 2 + 3 \times 100 \times 2^2 + 2^3
    • 1,000,000+60,000+1,200+8=1,061,2081,000,000 + 60,000 + 1,200 + 8 = 1,061,208
  • Problem b: Calculate 97297^2.
    • (100−3)2=1002−2×100×3+32(100 - 3)^2 = 100^2 - 2 \times 100 \times 3 + 3^2
    • 10,000−600+9=9,40910,000 - 600 + 9 = 9,409
  • Problem c: Calculate 12412^4.
    • (10+2)4=104+4×103×2+6×102×22+4×10×23+24(10 + 2)^4 = 10^4 + 4 \times 10^3 \times 2 + 6 \times 10^2 \times 2^2 + 4 \times 10 \times 2^3 + 2^4
    • 10,000+8,000+2,400+320+16=20,73610,000 + 8,000 + 2,400 + 320 + 16 = 20,736
  • Problem d: Calculate 1052105^2.
    • (100+5)2=1002+2×100×5+52(100 + 5)^2 = 100^2 + 2 \times 100 \times 5 + 5^2
    • 10,000+1,000+25=11,02510,000 + 1,000 + 25 = 11,025

Aufgabe 6: Vollständige Induktion (Complete Induction)

  • Goal: Prove that for n∈Nn ∈ ℕ, the following sum of squares identity holds:
    • S(n)=∑k=1nk2=16n(n+1)(2n+1)S(n) = \frac{\sum}{k=1}^n k^2 = \frac{1}{6}n(n+1)(2n+1) (*)
  • Induktionsanfang (Base Case):
    • For n=1n = 1:
    • Left side: ∑k=11k2=1\frac{\sum}{k=1}^1 k^2 = 1
    • Right side: 16×1×(1+1)×(2×1+1)=16×1×2×3=1\frac{1}{6} \times 1 \times (1+1) \times (2 \times 1 + 1) = \frac{1}{6} \times 1 \times 2 \times 3 = 1
    • The statement holds for n=1n = 1.
  • Induktionsbehauptung (Inductive Hypothesis):
    • Assume equation (*) holds for an arbitrary value n∈Nn ∈ ℕ.
  • Induktionsschritt (Inductive Step):
    • Show it holds for n+1n+1:
    • S(n+1)=∑k=1n+1k2=S(n)+(n+1)2S(n+1) = \frac{\sum}{k=1}^{n+1} k^2 = S(n) + (n+1)^2
    • Substitute the hypothesis for S(n)S(n):
    • =16n(n+1)(2n+1)+(n+1)2= \frac{1}{6}n(n+1)(2n+1) + (n+1)^2
    • Factor out (n+1)(n+1):
    • =16(n+1)[n(2n+1)+6(n+1)]= \frac{1}{6}(n+1) [n(2n+1) + 6(n+1)]
    • Expand within the brackets:
    • =16(n+1)[2n2+n+6n+6]=16(n+1)[2n2+7n+6]= \frac{1}{6}(n+1) [2n^2 + n + 6n + 6] = \frac{1}{6}(n+1) [2n^2 + 7n + 6]
    • Factor the quadratic expression 2n2+7n+62n^2 + 7n + 6 into (n+2)(2n+3)(n+2)(2n+3):
    • =16(n+1)(n+2)(2n+3)= \frac{1}{6}(n+1)(n+2)(2n+3)
    • Rewrite to match the target form:
    • =16(n+1)((n+1)+1)(2(n+1)+1)= \frac{1}{6}(n+1)((n+1)+1)(2(n+1)+1)
    • This is exactly the expression (*) for n+1n+1. Q.E.D.