To prove Proprieties P1, P2, P3, and P7 of Theorem 3 , let X = log a M and Y = log a N , and give reasons for the steps listed in Exercises 119 – 122. Proof of P2 of Theorem 3. M = a X and N = a Y , definition of logarithm So M N = a X a Y = a X − Y Quotient Rule for exponents Thus, log a M N = X − Y Definition of logarithm = log a M + log a N substitution
To prove Proprieties P1, P2, P3, and P7 of Theorem 3 , let X = log a M and Y = log a N , and give reasons for the steps listed in Exercises 119 – 122. Proof of P2 of Theorem 3. M = a X and N = a Y , definition of logarithm So M N = a X a Y = a X − Y Quotient Rule for exponents Thus, log a M N = X − Y Definition of logarithm = log a M + log a N substitution
Solution Summary: The author explains the reasons behind each step for the proof of the logarithmic property mathrmlog_a(MN)=
Taylor Series Approximation Example- H.W
More terms used implies better approximation
f(x) 4
f(x)
Zero order
f(x + 1) = f(x;)
First order
f(x; + 1) = f(x;) + f'(x;)h
1.0
Second order
0.5
True
f(x + 1) =
f(x) + f'(x)h +
ƒ"(x;)
h2
2!
f(x+1)
0
x; = 0
x+1 = 1
x
h
f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2
51
Taylor Series Approximation H.w:
Smaller step size implies smaller error
Errors
f(x) +
f(x,)
Zero order
f(x,+ 1) = f(x)
First order
1.0
0.5
Reduced step size
Second order
True
f(x + 1) = f(x) + f'(x)h
f(x; + 1) = f(x) + f'(x)h + "(xi) h2
f(x,+1)
O
x₁ = 0
x+1=1
Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5
iterations (or & s= 5%) for
f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2
52
Could you explain this using the formula I attached and polar coorindates
Could you explain this using the formula I attached and polar coordinates
Chapter 3 Solutions
Calculus and Its Applications Plus MyLab Math with Pearson eText -- Access Card Package (11th Edition) (Bittinger, Ellenbogen & Surgent, The Calculus and Its Applications Series)
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