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I have trouble understanding the following problem:

"Consider the function $f(x)=(5-2x)^{-1}$: Proof by induction, for all $n \in \mathbb{N}_0$: $$f^{(n)}(x) = 2^n n!(5-2x)^{-(n+1)}$$ Additionally calclulate the taylorseries of $f$ for $x_0 = 1$".

What exactly should be proofed here? I have trouble understanding the notation. For wich $f$ should the Taylor series be calculated?

user13764911
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3 Answers3

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You have to show that the $n-$th derivative of $f(x)$ has that required form using induction. That means, you show that the first derivative is $f'(x)=2^1\cdot n!\cdot (5-2x)^{-(1+1)}$ and that if the $n-$th derivative of $f$ is given by the formula, then $f^{(n+1)}(x)$ has the desired formula. So, essentially, you assumed the formula for $f^{(n)}(x)$ and show it for $f^{(n+1)}(x).$ For that, just differentiate once.

Remember that taylor expansion around a point $x_0$ is $$f(x)=\sum _{n=0}^{\infty}f^{(n)}(x_0)\frac{(x-x_0)^n}{n!},$$ so you use the formula you showed at $x_0=1.$

Phicar
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Welcome to MSE. If you try some first derivation you will find the clue. $$f(x)=(5-2x)^{-1}\\f'=(-1)(5-2x)^{-2}(-2)=+2(5-2x)^{-2}(1)\\ f''=+2(-2)(5-2x)^{-3}(-2)=+2^2(5-2x)^{-3}(2.1)\\f'''=2^2(-3)(5-2x)^{-4}(-2)=2^3(5-2x)^{-4}(3.2.1)\\f^4=2^4(5-2x)^{-5}(4.3.2.1)\\\vdots$$ $$f^{(n)}(x) = 2^n n!(5-2x)^{-(n+1)}\\n\to n+1\\(f^{(n)}(x))' =( 2^n n!(5-2x)^{-(n+1)})'\\= 2^n n!(5-2x)^{-(n+2)}(-2)(-(n+1))\\=2^{n+1}(n!(n+1))(5-2x)^{-(n+2)}$$

Khosrotash
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For Taylor series arround $ x_0=1$.

Let $$g(x)=f(x+1)=\frac{1}{3-2x}=\frac13\frac{1}{1-\frac 23 x}$$ $$=\frac 13\frac{1}{1-X}$$ $$=\frac 13\sum_{n=0}^NX^i+o(X^N)$$

With $ X=\frac 23x $ .

So $$f(x)=$$ $$\frac 13\sum_{n=0}^N(\frac 23)^n(x-1)^n+o((x-1)^N)$$

hamam_Abdallah
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