Math, asked by shivambishal192, 4 days ago


 {e}^{i\pi}  =  - 1
why this happens?​

Answers

Answered by mathdude500
7

\large\underline{\sf{Solution-}}

In order to prove the result, let we first derive Eulerian form.

That means, let we first prove,

\rm \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \: {e}^{ix}  \:  =  \: cosx \:  +  \: i \: sinx \\

We know,

\rm \:  {e}^{x} \:  =  \: 1 + x + \dfrac{ {x}^{2} }{2!}  + \dfrac{ {x}^{3} }{3!} +\dfrac{ {x}^{4} }{4!}  -  -  -  -  \\

\rm \:  \:  \:  \:  \:  \:  \:  \: Replace \: x \: by \: ix , \: we \: get\\

\rm \:  {e}^{ix} \:  =  \: 1 + ix + \dfrac{ { {i}^{2} x}^{2} }{2!}  + \dfrac{  {i}^{3} {x}^{3} }{3!} +\dfrac{ {i}^{4}  {x}^{4} }{4!}  -  -  -  -  \\

We know,

\boxed{\sf{  \: {i}^{2} =  - 1 \: }} \\  \\ \boxed{\sf{  \: {i}^{3}  =  - i \: }} \\  \\ \boxed{\sf{  \: {i}^{4}  = 1 \:  \: }} \\

So, on substituting these values, we get

\rm \:  {e}^{ix} \:  =  \: 1 + ix  -  \dfrac{ {x}^{2} }{2!}  -  \dfrac{i {x}^{3} }{3!} +\dfrac{ {x}^{4} }{4!}  -  -  -  -  \\

can be re-arranged as

\rm \:  {e}^{ix} = \bigg(1 - \dfrac{ {x}^{2} }{2!}  + \dfrac{ {x}^{4} }{4!} +  -  -  - \bigg) + i\bigg(x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} +  -  -  - \bigg)  \\

We know,

\boxed{\sf{  \:\rm \:  cosx - \dfrac{ {x}^{2} }{2!}  + \dfrac{ {x}^{4} }{4!} +  -  -  -  \:  \:  \: }}  \\  \\ \boxed{\sf{  \:\rm \: sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} +  -  -  -}}   \\

So, on substituting the values, we get

\rm\implies \: {e}^{ix} = cosx + isinx \\

Now,

\rm \:  {e}^{i\pi}  \\

\rm \:  \:  \:  \:  \:  \:  =  \: cos\pi \:  +  \: i \: sin\pi \\

\rm \:  \:  \:  \:  \:  \:  =   - 1 + 0i \\

\rm \:  \:  \:  \:  \:  \:  =   - 1 \\

Hence,

\rm\implies \: {e}^{i\pi}  \:  =  \:  -  \: 1 \:  \\

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

Some important expansions :-

\rm \: sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} +  -  -  -  \\

\rm \: cosx = 1 + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} +  -  -  -  \\

\rm \: tanx = x  -  \dfrac{ {x}^{3} }{3} + \dfrac{2 {x}^{5} }{15} +  -  -  -  \\

\rm \:  {e}^{ - x} \:  =  \: 1  -  x + \dfrac{ {x}^{2} }{2!}   -  \dfrac{ {x}^{3} }{3!} +\dfrac{ {x}^{4} }{4!}  -  -  -  -  \\

\rm \:  {a}^{ - x} \:  =  \: 1  -  xloga + \dfrac{ {x}^{2} }{2!} {(loga)}^{2}    -  \dfrac{ {x}^{3} }{3!} {(loga)}^{3}  +\dfrac{ {x}^{4} }{4!} {(loga)}^{4}   -  -  -  -  \\

\rm \:  {a}^{x} \:  =  \: 1 +  xloga + \dfrac{ {x}^{2} }{2!} {(loga)}^{2} +  \dfrac{ {x}^{3} }{3!} {(loga)}^{3}  +\dfrac{ {x}^{4} }{4!} {(loga)}^{4}   -  -  -  -  \\

Answered by IamOnePunchMan
4

\large\underline{\sf{Solution-}}

In order to prove the result, let we first derive Eulerian form.

That means, let we first prove,

\rm \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \: {e}^{ix}  \:  =  \: cosx \:  +  \: i \: sinx \\

We know,

\rm \:  {e}^{x} \:  =  \: 1 + x + \dfrac{ {x}^{2} }{2!}  + \dfrac{ {x}^{3} }{3!} +\dfrac{ {x}^{4} }{4!}  -  -  -  -  \\

\rm \:  \:  \:  \:  \:  \:  \:  \: Replace \: x \: by \: ix , \: we \: get\\

\rm \:  {e}^{ix} \:  =  \: 1 + ix + \dfrac{ { {i}^{2} x}^{2} }{2!}  + \dfrac{  {i}^{3} {x}^{3} }{3!} +\dfrac{ {i}^{4}  {x}^{4} }{4!}  -  -  -  -  \\

We know,

\boxed{\sf{  \: {i}^{2} =  - 1 \: }} \\  \\ \boxed{\sf{  \: {i}^{3}  =  - i \: }} \\  \\ \boxed{\sf{  \: {i}^{4}  = 1 \:  \: }} \\

So, on substituting these values, we get

\rm \:  {e}^{ix} \:  =  \: 1 + ix  -  \dfrac{ {x}^{2} }{2!}  -  \dfrac{i {x}^{3} }{3!} +\dfrac{ {x}^{4} }{4!}  -  -  -  -  \\

can be re-arranged as

\rm \:  {e}^{ix} = \bigg(1 - \dfrac{ {x}^{2} }{2!}  + \dfrac{ {x}^{4} }{4!} +  -  -  - \bigg) + i\bigg(x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} +  -  -  - \bigg)  \\

We know,

\boxed{\sf{  \:\rm \:  cosx - \dfrac{ {x}^{2} }{2!}  + \dfrac{ {x}^{4} }{4!} +  -  -  -  \:  \:  \: }}  \\  \\ \boxed{\sf{  \:\rm \: sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} +  -  -  -}}   \\

So, on substituting the values, we get

\rm\implies \: {e}^{ix} = cosx + isinx \\

Now,

\rm \:  {e}^{i\pi}  \\

\rm \:  \:  \:  \:  \:  \:  =  \: cos\pi \:  +  \: i \: sin\pi \\

\rm \:  \:  \:  \:  \:  \:  =   - 1 + 0i \\

\rm \:  \:  \:  \:  \:  \:  =   - 1 \\

Hence,

\rm\implies \: {e}^{i\pi}  \:  =  \:  -  \: 1 \:  \\

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

Some important expansions :-

\rm \: sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} +  -  -  -  \\

\rm \: cosx = 1 + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} +  -  -  -  \\

\rm \: tanx = x  -  \dfrac{ {x}^{3} }{3} + \dfrac{2 {x}^{5} }{15} +  -  -  -  \\

\rm \:  {e}^{ - x} \:  =  \: 1  -  x + \dfrac{ {x}^{2} }{2!}   -  \dfrac{ {x}^{3} }{3!} +\dfrac{ {x}^{4} }{4!}  -  -  -  -  \\

\rm \:  {a}^{ - x} \:  =  \: 1  -  xloga + \dfrac{ {x}^{2} }{2!} {(loga)}^{2}    -  \dfrac{ {x}^{3} }{3!} {(loga)}^{3}  +\dfrac{ {x}^{4} }{4!} {(loga)}^{4}   -  -  -  -  \\

\rm \:  {a}^{x} \:  =  \: 1 +  xloga + \dfrac{ {x}^{2} }{2!} {(loga)}^{2} +  \dfrac{ {x}^{3} }{3!} {(loga)}^{3}  +\dfrac{ {x}^{4} }{4!} {(loga)}^{4}   -  -  -  -  \\

Similar questions