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Reil [10]
2 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]2 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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Step-by-step explanation:

13*9 = $117

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$117 + $33 = $150

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3 years ago
Solve for x.<br> X + 10<br> 2x - 50<br><br> x = [?]
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Answer:

x +10 = 2x-50

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2 years ago
Which fractions 1/3, 5/6, 3/4, 3/8 are closer to 0 than to 1?
aev [14]

Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
Determine the equation of the polynomial, p(x),in both factored and standard form.
melamori03 [73]

We can see that the graph touches x=-1 without crossing the x-axis (i.e. it is a double solution), and then there's another zero at x=2 (this time it's a crossing zero, so a single solution).

This leads, up to multiple, to the polynomial

p(x)=a(x+1)^2(x-2)

If we impose the passing through (0,4) we have

p(0)=4=a(1)(-2) \iff -2a=4 \iff a=-2

So, the polynomial is

p(x)=-2(x+1)^2(x-2)=-2 x^3 + 6 x + 4

Finally, to solve p(x), simply look at the graph, searching for the points, where the graph is below the x-axis. You can see that this happens only if x>2, so that's the solution to your question.

4 0
3 years ago
How should the decimal point in 34.05 be moved to determine the product 34.05 × 10 to the power of 6?
Ray Of Light [21]

Answer:

The decimal point should be moved Six places to the right side because there are Six zeros in 10 to the power of 6.

Step-by-step explanation:

As we have to represent the number in the form of 10^{6} we need to move the decimal Six places towards left because there are Six zeros in 10^{6} i.e. 1000000. For this purpose we will add four zeros to the left of the number.

So the number in the format mentioned in question will look like -

0.00003405 × 10^{6}.

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3 years ago
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