[tex]\begin{gathered} \rm Prove \: that :- \\ { { \rm { f ( x ) = f ( 0 ) + \dfrac{x}{1!} f'(0) + \dfrac{x²}{2!} f''(0) + \dfrac{x³}{3!} f'''(0) + \cdot \cdot \cdot \cdot \cdot \cdot \cdot }}}\end{gathered}[/tex]




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The series you had provided is generally the Maclaurin's series , and we want to prove the series , this series is very very much useful in calculus , especially when you have to substitute a function by it's series . So now let's start !

Consider the function f(x) as series of a polynomials of nth (n ≥ 1) degree as

[tex]{\boxed{\bf{f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \cdots \infty}}}[/tex]

Now , at x = 0

[tex]{:\implies \quad \sf f(0)=a_{0}+a_{1}(0)+a_{2}(0)^{2}+a_{3}(0)^{3}+a_{4}(0)^{4}+\cdots \cdots \infty}[/tex]

[tex]{\boxed{:\implies \quad \sf f(0)=a_{0}}}[/tex]

Now , again consider the function that we assumed

[tex]{:\implies \quad \sf f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \cdots \infty}[/tex]

Differentiating both sides w.r.t.x will yield

[tex]{:\implies \quad \sf f^{\prime}(x)=0+a_{1}\times 1+a_{2}\times 2\times x+a_{3}\times 3\times x^{2}+a_{4}\times 4\times x^{3}+\cdots \cdots \infty}[/tex]

[tex]{:\implies \quad \sf f^{\prime}(x)=a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}\cdots \cdots \infty}[/tex]

Now , put x = 0 ;

[tex]{:\implies \quad \sf f^{\prime}(0)=a_{1}+2a_{2}(0)+3a_{3}(0)^{2}+4a_{4}(0)^{3}\cdots \cdots \infty}[/tex]

[tex]{:\implies \quad \sf f^{\prime}(0)=a_{1}}[/tex]

[tex]{\boxed{:\implies \quad \sf a_{1}=\dfrac{f^{\prime}(0)}{1!}}}[/tex]

Now , consider ;

[tex]{:\implies \quad \sf f^{\prime}(x)=a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}\cdots \cdots \infty}[/tex]

Differentiating both sides w.r.t.x now ;

[tex]{:\implies \quad \sf f^{\prime \prime}(x)=0+2\times a_{2}\times 1+3\times a_{3}\times 2\times x+4\times a_{4}\times 3\times x^{2}\cdots \cdots \infty}[/tex]

[tex]{:\implies \quad \sf f^{\prime \prime}(x)=2a_{2}+6a_{3}x+12a_{4}x^{2}+\cdots \cdots \infty}[/tex]

Now , putting x = 0 will yield ;

[tex]{:\implies \quad \sf f^{\prime \prime}(0)=2a_{2}+6a_{3}(0)+12a_{4}(0)^{2}+\cdots \cdots \infty}[/tex]

[tex]{:\implies \quad \sf f^{\prime \prime}(0)=2a_{2}}[/tex]

[tex]{\boxed{:\implies \quad \sf a_{2}=\dfrac{f^{\prime \prime}(0)}{2!}}}[/tex]

Now , similarly we can prove that ;

[tex]{\boxed{:\implies \quad \sf a_{n}=\dfrac{f^{n}(0)}{n!}}}[/tex]

Now , as we had considered ;

[tex]{:\implies \quad \sf f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \cdots \infty}[/tex]

Now , putting the values we obtained above ;

[tex]{:\implies \quad \sf f(x)=f(0)+\dfrac{x}{1!}f^{\prime}(0)+\dfrac{x^2}{2!}f^{\prime \prime}(0)+\dfrac{x^3}{3!}f^{\prime \prime \prime}(0)+\dfrac{x^4}{4!}f^{\prime \prime \prime \prime}(0)+\cdots \cdots \dfrac{x^n}{n!}f^{n}(0)+\cdots \cdots \infty}[/tex]

Hence , Proved

solve for x ~

[tex]x {}^{2} - 5x + 6 = 0[/tex]

thankyou ~​

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