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bulgar [2K]
3 years ago
10

Solve for x: −2(x + 3) = −2(x + 1) − 4

Mathematics
2 answers:
stira [4]3 years ago
6 0
It is D no solution I think
mash [69]3 years ago
4 0
-2(x + 3) = -2(x + 1) - 4
-2x - 6 = -2x - 2 - 4
0x = -6 + 6
0x = 0

x = unknown
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Consider the function ​f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0
dybincka [34]

I suppose you mean

f(x)=\cos(x^2)

Recall that

\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}

which converges everywhere. Then by substitution,

\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}

(starting at n=1 because the summand is 0 when n=0)

b. Naturally, the differentiated series represents

f'(x)=-2x\sin(x^2)

To see this, recalling the series for \sin x, we know

\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}

Multiplying by -2x gives

-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}

and from here,

-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}

-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)

c. This series also converges everywhere. By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}

The limit is 0, so any choice of x satisfies the convergence condition.

3 0
3 years ago
What could this be???????
Step2247 [10]
Your answer is A it's a pretty simple task that you may just be overthinking
8 0
3 years ago
Read 2 more answers
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