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Reil [10]
3 years ago
8

Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that

Rn(x)→0 . f(x)=lnx, a=
Mathematics
1 answer:
Kobotan [32]3 years ago
3 0

Answer:

Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).

Remember that the general Taylor expansion is:

f(x) = f(a) + f'(a)*(x - a) + \frac{1}{2!}*f''(a)(x -a)^2 + ...

for our function we have:

f'(x) =  1/x

f''(x) = -1/x^2

f'''(x) =  (1/2)*(1/x^3)

this is enough, now just let's write the series:

f(x) = ln(a) +  \frac{1}{a} *(x - a) - \frac{1}{2!} *\frac{1}{a^2} *(x - a)^2 + \frac{1}{3!} *\frac{1}{2*a^3} *(x - a)^3 + ....

This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.

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Answer:

The correct answer is  option C

They are complementary angles

Step-by-step explanation:

From the figure we can see two angles <ABC and <LMN,

<u>To find the correct option</u>

From figure we get,

m<ABC = 20°° and m<LMN = 70°

m<ABC + m<LMN = 20 + 70 = 90°

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The correct answer is option C

They are complementary angles

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What’s the answer for this?
Yuliya22 [10]

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5 0
3 years ago
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Step-by-step explanation:

Given bi-quadratic equation is:

x^4+95x^2 -500=0

Substituting x^2=a, given bi-quadratic equation reduces in the form of following quadratic equation:

a^2 +95a -500=0\\

Let us factorize the above quadratic equation:

a^2 +95a -500=0\\\therefore\: a^2 +100a -5a-500=0\\\therefore\: a(a +100) -5(a+100)=0\\\therefore\: (a +100)(a -5)=0\\\therefore\: (a +100)=0\:\: or \:\: (a -5)=0\\\therefore\: a = - 100\:\: or \:\: a = 5\\\\CASE\: (1)\:\\When\: a = - 100 \implies x^2 = - 100\\\therefore x=\pm \sqrt{-100}\\

Since square root of a negative number cannot be found, so:

x \neq \pm \sqrt{-100}\\

CASE\: (2)\:\\When\: a =5 \implies x^2 =5\\\therefore x=\pm \sqrt{5}\\

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3 years ago
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