Analiz (Calculus) - İspatlar ve Detaylı Anlatım

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Limit ve Süreklilik

Limits describe the behavior of functions as inputs approach a certain value. Continuity ensures functions have no jumps or breaks.

Limit Tanımı (ε-δFormal definition of limits using epsilon and delta Tanımı)

Let f: D → ℝ be a function and c be a limit point of D. We say the limit of f(x) as x approaches c is L, written as:

lim_{x→c} f(x) = L

if for every ε > 0, there exists a δ > 0 such that for all x ∈ D with 0 < |x - c| < δ, we have |f(x) - L| < ε.

Sandwich Theorem (Squeeze Theorem) Proof:

Suppose g(x) ≤ f(x) ≤ h(x) for all x in some interval containing c (except possibly at c), and:

lim_{x→c} g(x) = lim_{x→c} h(x) = L

Then lim_{x→c} f(x) = L.

Proof: For any ε > 0, choose δ₁, δ₂ > 0 such that:

If 0 < |x-c| < δ₁, then |g(x)-L| < ε ⇒ L-ε < g(x)

If 0 < |x-c| < δ₂, then |h(x)-L| < ε ⇒ h(x) < L+ε

Let δ = min(δ₁, δ₂). Then for 0 < |x-c| < δ:

L-ε < g(x) ≤ f(x) ≤ h(x) < L+ε ⇒ |f(x)-L| < ε

Thus lim_{x→c} f(x) = L. ∎

Süreklilik Tanımı

A function f is continuous at c if:

1. f(c) is defined
2. lim_{x→c} f(x) exists
3. lim_{x→c} f(x) = f(c)

Intermediate Value Theorem: If f is continuous on [a, b] and k is any number between f(a) and f(b), then there exists at least one c ∈ (a, b) such that f(c) = k.

Örnek: Limit Hesaplama

Evaluate lim_{x→0} (sin x)/x.

Using geometric proof and squeeze theorem:

For 0 < x < π/2: sin x < x < tan x

Divide by sin x: 1 < x/(sin x) < 1/(cos x)

Take reciprocals: cos x < (sin x)/x < 1

As x → 0, cos x → 1

By squeeze theorem: lim_{x→0} (sin x)/x = 1

Result: 1

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Türev ve Uygulamaları

The derivative measures the instantaneous rate of change of a function. It's fundamental to differential calculus.

Türev Tanımı

The derivative of f at x is:

f'(x) = lim_{h→0} [f(x+h) - f(x)]/h

provided this limit exists. Geometrically, it represents the slope of the tangent line.

Türev Kuralları

Power Rule

d/dx [xⁿ] = n·xⁿ⁻¹

Product Rule

(fg)' = f'g + fg'

Quotient Rule

(f/g)' = (f'g - fg')/g²

Chain Rule

(f∘g)'(x) = f'(g(x))·g'(x)

Product Rule Proof:

Let F(x) = f(x)g(x). Then:

F'(x) = lim_{h→0} [F(x+h) - F(x)]/h = lim_{h→0} [f(x+h)g(x+h) - f(x)g(x)]/h

Add and subtract f(x+h)g(x):

= lim_{h→0} [f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)]/h = lim_{h→0} f(x+h)·[g(x+h)-g(x)]/h + lim_{h→0} g(x)·[f(x+h)-f(x)]/h = f(x)g'(x) + g(x)f'(x)

assuming f is continuous at x. ∎

Mean Value Theorem

Mean Value Theorem (MVTMean Value Theorem connects derivatives to average rates of change): If f is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that:

f'(c) = [f(b) - f(a)]/(b - a)

Mean Value Theorem Proof:

Define g(x) = f(x) - f(a) - [(f(b)-f(a))/(b-a)](x-a).

Note that g(a) = g(b) = 0. By Rolle's Theorem, there exists c ∈ (a, b) such that g'(c) = 0.

g'(x) = f'(x) - [f(b)-f(a)]/(b-a)

So g'(c) = 0 ⇒ f'(c) = [f(b)-f(a)]/(b-a). ∎

Örnek: Türev Uygulaması

Find the maximum area of a rectangle inscribed in a semicircle of radius r.

Let rectangle width = 2x, height = y

From semicircle equation: x² + y² = r² ⇒ y = √(r² - x²)

Area A = 2x·y = 2x√(r² - x²)

A'(x) = 2√(r² - x²) - 2x²/√(r² - x²) = 2(r² - 2x²)/√(r² - x²)

Set A'(x) = 0 ⇒ r² - 2x² = 0 ⇒ x = r/√2

Then y = r/√2, maximum area A_max = r²

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İntegral ve Uygulamaları

Integration is the reverse process of differentiation, used to calculate areas, volumes, and accumulated quantities.

RiemannBernhard Riemann, German mathematician who formalized integration Integral Tanımı

Let f be bounded on [a, b]. A partition P divides [a, b] into subintervals. The Riemann sum is:

R(f, P) = Σ_{i=1}^n f(t_i)Δx_i

where t_i ∈ [x_{i-1}, x_i]. f is Riemann integrable if these sums converge to a limit as the partition norm approaches 0.

Fundamental Theorem of Calculus:

Part 1: If f is continuous on [a, b], then F(x) = ∫_a^x f(t)dt is differentiable on (a, b) and F'(x) = f(x).

Part 2: If F is any antiderivative of f on [a, b], then:

∫_a^b f(x)dx = F(b) - F(a)

FTCFundamental Theorem of Calculus links derivatives and integrals Part 1 Proof:

For h > 0:

[F(x+h) - F(x)]/h = [1/h] ∫_x^{x+h} f(t)dt

By the Mean Value Theorem for Integrals, there exists c_h ∈ [x, x+h] such that:

∫_x^{x+h} f(t)dt = f(c_h)·h

Thus [F(x+h)-F(x)]/h = f(c_h). As h → 0, c_h → x, and by continuity of f, f(c_h) → f(x). So F'(x) = f(x). ∎

İntegral Teknikleri

Substitution Rule

∫ f(g(x))g'(x)dx = ∫ f(u)du

where u = g(x)

Integration by Parts

∫ u dv = uv - ∫ v du

Derived from product rule

Trigonometric Integrals

Using identities like sin²x + cos²x = 1

Partial Fractions

Decomposing rational functions

Örnek: İntegral Hesaplama

Evaluate ∫ x·e^x dx using integration by parts.

Let u = x, dv = e^x dx

Then du = dx, v = e^x

∫ u dv = uv - ∫ v du

∫ x e^x dx = x e^x - ∫ e^x dx

= x e^x - e^x + C

= e^x(x - 1) + C

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Diziler ve Seriler

Sequences are ordered lists of numbers, series are sums of sequences. Convergence tests determine if series approach finite limits.

Sequence Convergence

A sequence {a_n} converges to L if for every ε > 0, there exists N ∈ ℕ such that for all n ≥ N, |a_n - L| < ε.

lim_{n→∞} a_n = L

Series Convergence Tests

Divergence Test

If lim a_n ≠ 0, then Σ a_n diverges.

Integral Test

If f is positive, continuous, decreasing on [1, ∞), then Σ f(n) and ∫_1^∞ f(x)dx both converge or diverge.

Ratio Test

lim |a_{n+1}/a_n| = L

L < 1: converges; L > 1: diverges; L = 1: inconclusive

Root Test

lim |a_n|^{1/n} = L

Same convergence conditions as ratio test

Integral Test Proof Sketch:

For a decreasing function f on [1, ∞):

f(k+1) ≤ ∫_k^{k+1} f(x)dx ≤ f(k)

Summing from k = 1 to n-1:

Σ_{k=2}^n f(k) ≤ ∫_1^n f(x)dx ≤ Σ_{k=1}^{n-1} f(k)

Thus the sequence of partial sums and the integral either both converge or both diverge. ∎

Power Series

A power series centered at c is:

Σ_{n=0}^∞ a_n (x-c)^n

It converges within an interval of radius R (radius of convergence).

Taylor's Theorem: If f has n+1 continuous derivatives on an interval containing a, then for x in that interval:

f(x) = Σ_{k=0}^n [f^{(k)}(a)/k!] (x-a)^k + R_n(x)

where the remainder R_n(x) = [f^{(n+1)}(c)/(n+1)!] (x-a)^{n+1} for some c between a and x.

Örnek: Seri Yakınsaklık Testi

Test convergence of Σ_{n=1}^∞ 1/n².

Use integral test with f(x) = 1/x²

∫_1^∞ 1/x² dx = lim_{b→∞} [-1/x]_1^b = lim_{b→∞} (1 - 1/b) = 1

The improper integral converges to 1

Therefore, by the integral test, the series converges

This is the famous Basel problem: Σ 1/n² = π²/6

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Çok Değişkenli Fonksiyonlar

Functions of several variables extend calculus concepts to higher dimensions, with partial derivatives and multiple integrals.

Partial Derivatives

The partial derivative of f(x, y) with respect to x is:

∂f/∂x = lim_{h→0} [f(x+h, y) - f(x, y)]/h

keeping y constant. Similarly for ∂f/∂y.

Clairaut's Theorem: If f has continuous second partial derivatives on an open set, then:

∂²f/∂x∂y = ∂²f/∂y∂x

The order of differentiation doesn't matter.

Gradient and Directional Derivative

The gradient of f: ℝⁿ → ℝ is:

∇f = (∂f/∂x₁, ∂f/∂x₂, ..., ∂f/∂xₙ)

The directional derivative in direction u (unit vector) is:

D_u f = ∇f · u

Chain Rule for Multivariable Functions:

If z = f(x, y) where x = g(t) and y = h(t) are differentiable, then:

dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt)

Proof: Using linear approximation:

Δz ≈ (∂f/∂x)Δx + (∂f/∂y)Δy

Divide by Δt and take limit as Δt → 0:

dz/dt = lim_{Δt→0} Δz/Δt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt). ∎

Multiple Integrals

The double integral of f(x, y) over region R is:

∬_R f(x, y) dA = lim_{max ΔA_i → 0} Σ f(x_i*, y_i*) ΔA_i

Evaluated as iterated integrals: ∫ [∫ f(x, y) dy] dx.

Örnek: Kısmi Türev

Find all second partial derivatives of f(x, y) = x³y + e^{xy}.

∂f/∂x = 3x²y + ye^{xy}

∂f/∂y = x³ + xe^{xy}

∂²f/∂x² = 6xy + y²e^{xy}

∂²f/∂y² = x²e^{xy}

∂²f/∂x∂y = 3x² + e^{xy} + xye^{xy}

∂²f/∂y∂x = 3x² + e^{xy} + xye^{xy} (same as above, confirming Clairaut)

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Diferansiyel Denklemler

Differential equations describe relationships involving rates of change. They model phenomena in physics, engineering, and biology.

Temel Tanımlar

An ordinary differential equation (ODEOrdinary Differential Equation involves one independent variable) relates a function and its derivatives:

F(x, y, y', y'', ..., y⁽ⁿ⁾) = 0

The order is the highest derivative present. A solution is a function that satisfies the equation.

First Order ODEOrdinary Differential Equation involves one independent variables

Separable Equations

dy/dx = f(x)g(y)

Solution: ∫ dy/g(y) = ∫ f(x)dx

Linear Equations

dy/dx + P(x)y = Q(x)

Solved using integrating factor μ(x) = e^{∫P(x)dx}

Exact Equations

M(x,y)dx + N(x,y)dy = 0

where ∂M/∂y = ∂N/∂x

Solution of Linear First Order ODEOrdinary Differential Equation involves one independent variable:

For y' + P(x)y = Q(x), multiply by integrating factor μ(x) = e^{∫P(x)dx}:

μ(x)y' + μ(x)P(x)y = μ(x)Q(x)

Note that (μ(x)y)' = μ(x)y' + μ'(x)y = μ(x)y' + μ(x)P(x)y since μ' = μP.

Thus: (μy)' = μQ

Integrate: μ(x)y = ∫ μ(x)Q(x)dx + C

So: y = [1/μ(x)] [∫ μ(x)Q(x)dx + C]. ∎

Second Order Linear ODEOrdinary Differential Equation involves one independent variables

For ay'' + by' + cy = 0 (homogeneous), the characteristic equation is:

ar² + br + c = 0

Roots r₁, r₂ determine solution form:

Real distinct: y = C₁e^{r₁x} + C₂e^{r₂x}
Real equal: y = (C₁ + C₂x)e^{rx}
Complex: y = e^{αx}(C₁cosβx + C₂sinβx)

Laplace Transforms

The Laplace transform converts differential equations to algebraic equations:

L{f(t)} = F(s) = ∫_0^∞ e^{-st} f(t) dt

Useful for solving ODEOrdinary Differential Equation involves one independent variables with initial conditions.

Örnek: Diferansiyel Denklem Çözümü

Solve y' + 2y = e^{-x}, y(0) = 1.

This is linear: P(x) = 2, Q(x) = e^{-x}

Integrating factor: μ(x) = e^{∫2dx} = e^{2x}

Multiply: e^{2x}y' + 2e^{2x}y = e^{x}

Left side is (e^{2x}y)', so integrate: e^{2x}y = ∫ e^{x}dx = e^{x} + C

Thus: y = e^{-x} + Ce^{-2x}

Use y(0) = 1: 1 = 1 + C ⇒ C = 0

Solution: y = e^{-x}