Question

$\\ { \bold{ \underline{Prove \: \: that} : - }}$

$\\ \: \: \: \: \: { \bold{ {e}^{x} = \left( \dfrac{ { \triangle}^{2} }{E} \right) {e}^{x} \left \{ \dfrac{E( {e}^{x} )}{ { \triangle}^{2}. {e}^{x} } \right \} }}$

• Here –

$\\ \: \: \: \: { \huge{.}} { \bold{ \: \: \triangle \: = forward \: \: operator }}$

$\\ \: \: \: \: { \huge{.}} { \bold{ \: \: E \: = Displacement \: \: operator }}$

• Don't give irrelative answers.​

Answer 1

${\sf{\tt{\underline{Given:}}}}$

$\begin{gathered} \dagger \: \: { \sf{ {e}^{x} = \left( \dfrac{ { \triangle}^{2} }{E} \right) {e}^{x} .\left \{ \dfrac{E( {e}^{x} )}{ { \triangle}^{2}. {e}^{x} } \right \} }} \\ \end{gathered}$

${\bf{\rm{\underline{Now:}}}}$

$\begin{gathered} {\rm{ \sf \underline{ \: \green{ First \: Method : }}}} \\ \\ \end{gathered}$

Take R.H.S,

$\begin{gathered} \: { \sf{ = \left( \dfrac{ { \triangle}^{2} }{E} \right) {e}^{x} .\left \{ \dfrac{E( {e}^{x} )}{ { \triangle}^{2}. {e}^{x} } \right \} }} \\ \end{gathered}$

We know that,

$\begin{gathered} : \implies \boxed{\sf{ \triangle \equiv e - 1}} \\ \\ \end{gathered}$

$\begin{gathered} \: {\sf{ \bigg [ \frac{( {E - 1)}^{2} }{E} \bigg] {e}^{x} \frac{e \: {e}^{x} }{( {E - 1)}^{2} {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} : \implies \boxed{\sf{ ( {x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }} \\ \\ \end{gathered}$

$\begin{gathered} \: = {\sf{ \bigg( \frac{ {E}^{2} - 2E + 1}{E} \bigg) {e}^{x}. \frac{e \: {e}^{x} }{( {E}^{2} - 2E + 2) {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} \: = {\sf{ \bigg( \frac{ {E}^{2} }{e} - \frac{2E}{E} + \frac{1}{E} \bigg) {e}^{x} . \frac{E\: {e}^{x} }{( {E}^{2} - 2E+ 1) {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = (E - 2 + {E}^{ - 1} ) {e}^{x} . \frac{E \: {e}^{x} }{( {E}^{2} - 2E + 1) {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = E {e}^{x} - 2 {e}^{x} + {E}^{ - 1} {e}^{x} . \frac{E \: {e}^{x} }{{E}^{2} {e}^{x} - 2E {e}^{x} + 1{e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{x + 1} - 2 {e}^{x} + {e}^{x +1} . \frac{ \: {e}^{x + 1} }{ {e}^{x + 2} - 2{e}^{x + 1} + {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = \cancel{{e}^{x + 1} - 2 {e}^{x} + {e}^{x-1}} . \frac{ \: {e}^{x } }{ \cancel{{e}^{x + 1} - 2{e}^{x } + {e}^{x-1} } }}} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{x} }} \\ \\ \end{gathered}$

★ Hence L.H.S = R.H.S

____________________________

$\begin{gathered} {\tt{ \dagger \underline{ \: \green{ Second \: method : }}}} \\ \\ \end{gathered}$

Consider

$\begin{gathered} : \implies{\sf{ \frac{ \triangle^{2} }{E} {e}^{x} = \triangle^{2} {E}^{ - 1} {e}^{x} }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = \triangle^{2} {e}^{x}. {e}^{ - h} }} \\ \\ \end{gathered}$

Now R.H.S,

$\begin{gathered} {\sf{ \: = \triangle^{2} {e}^{x}. {e}^{ - h} . \frac{E \: {e}^{x} }{ { \triangle}^{2} {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = \cancel{\triangle^{2} {e}^{x}} {e}^{ - h} . \frac{E \: {e}^{x} }{ \cancel{ { \triangle}^{2} {e}^{x} } }}} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{ - h} e \: {e}^{x} }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{ - h} \: {e}^{x + h} }} \\ \\ \end{gathered}$

$\begin{gathered} {\tt{ \: = {e}^{x} }} \\ \end{gathered}$

Answer 2

$\begin{gathered} \dagger \: \: { \sf{ {e}^{x} = \left( \dfrac{ { \triangle}^{2} }{E} \right) {e}^{x} .\left \{ \dfrac{E( {e}^{x} )}{ { \triangle}^{2}. {e}^{x} } \right \} }} \\ \end{gathered}$

${\bf{\blue{\underline{Now:}}}}$

$\begin{gathered} {\frak{ \dagger \underline{ \: \green{ first \: method : }}}} \\ \\ \end{gathered}$

Take R.H.S,

$\begin{gathered} \: { \sf{ = \left( \dfrac{ { \triangle}^{2} }{E} \right) {e}^{x} .\left \{ \dfrac{E( {e}^{x} )}{ { \triangle}^{2}. {e}^{x} } \right \} }} \\ \end{gathered}$

We know that,

$\begin{gathered} : \implies \boxed{\sf{ \triangle \equiv e - 1}} \\ \\ \end{gathered}$

$\begin{gathered} \: {\sf{ \bigg [ \frac{( {E - 1)}^{2} }{E} \bigg] {e}^{x} \frac{e \: {e}^{x} }{( {E - 1)}^{2} {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} : \implies \boxed{\sf{ ( {x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }} \\ \\ \end{gathered}$

$\begin{gathered} \: = {\sf{ \bigg( \frac{ {E}^{2} - 2E + 1}{E} \bigg) {e}^{x}. \frac{e \: {e}^{x} }{( {E}^{2} - 2E + 2) {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} \: = {\sf{ \bigg( \frac{ {E}^{2} }{e} - \frac{2E}{E} + \frac{1}{E} \bigg) {e}^{x} . \frac{E\: {e}^{x} }{( {E}^{2} - 2E+ 1) {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = (E - 2 + {E}^{ - 1} ) {e}^{x} . \frac{E \: {e}^{x} }{( {E}^{2} - 2E + 1) {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = E {e}^{x} - 2 {e}^{x} + {E}^{ - 1} {e}^{x} . \frac{E \: {e}^{x} }{{E}^{2} {e}^{x} - 2E {e}^{x} + 1{e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{x + 1} - 2 {e}^{x} + {e}^{x +1} . \frac{ \: {e}^{x + 1} }{ {e}^{x + 2} - 2{e}^{x + 1} + {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = \cancel{{e}^{x + 1} - 2 {e}^{x} + {e}^{x-1}} . \frac{ \: {e}^{x } }{ \cancel{{e}^{x + 1} - 2{e}^{x } + {e}^{x-1} } }}} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{x} }} \\ \\ \end{gathered}$

Hence L.H.S = R.H.S

____________________________

$\begin{gathered} {\frak{ \dagger \underline{ \: \green{ Second \: method : }}}} \\ \\ \end{gathered}$

Consider

$\begin{gathered} : \implies{\sf{ \frac{ \triangle^{2} }{E} {e}^{x} = \triangle^{2} {E}^{ - 1} {e}^{x} }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = \triangle^{2} {e}^{x}. {e}^{ - h} }} \\ \\ \end{gathered}$

Now R.H.S,

$\begin{gathered} {\sf{ \: = \triangle^{2} {e}^{x}. {e}^{ - h} . \frac{E \: {e}^{x} }{ { \triangle}^{2} {e}^{x} } }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = \cancel{\triangle^{2} {e}^{x}} {e}^{ - h} . \frac{E \: {e}^{x} }{ \cancel{ { \triangle}^{2} {e}^{x} } }}} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{ - h} e \: {e}^{x} }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{ - h} \: {e}^{x + h} }} \\ \\ \end{gathered}$

$\begin{gathered} {\sf{ \: = {e}^{x} }} \\ \\ \end{gathered}$

Hence L.HS = R.H.S