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Simplification of a complicated fraction
The 2019 Stack Overflow Developer Survey Results Are InA question about how to express a fraction as $1over q_1+1over q_2+ cdots+1over q_N$Simplification imaginary fractionsElimating ODE constant of a extended surface problemOrthogonality between eigenfunctions of euler-bernoulli differential equation for a simple beamIntegral in physics - how to evaluate it?Kronig Penney Model Matrix DeterminantHow many times do you have go back and forth to get out of a deadlock?Calculation of capacitance between two cylindersFraction and simplificationHow could a square root of fraction have a negative root?
$begingroup$
I am going over a physics text and I have difficulty to see how one can go from
$$2A = (1+ fracalphaik)(1+fracikalpha)fracFe^ikae^-alpha a2 + (1- fracalphaik)(1-fracikalpha)fracFe^ikae^alpha a2$$
to
$$fracAe^-ikaF = cosh(alpha a) + i(fracalpha^2 - k^22k alpha)sinh(alpha a)$$
Could you give me some hints on how to proceed? I, obviously, know $cosh(x) = frace^x+e^-x2$ and $sinh(x) = frace^x-e^-x2$ but the connection between the two steps is not obvious to me.
physics fractions
asked Mar 23 at 5:58
RobRob
20219
$endgroup$
add a comment |
$begingroup$
I am going over a physics text and I have difficulty to see how one can go from
$$2A = (1+ fracalphaik)(1+fracikalpha)fracFe^ikae^-alpha a2 + (1- fracalphaik)(1-fracikalpha)fracFe^ikae^alpha a2$$
to
$$fracAe^-ikaF = cosh(alpha a) + i(fracalpha^2 - k^22k alpha)sinh(alpha a)$$
Could you give me some hints on how to proceed? I, obviously, know $cosh(x) = frace^x+e^-x2$ and $sinh(x) = frace^x-e^-x2$ but the connection between the two steps is not obvious to me.
physics fractions
asked Mar 23 at 5:58
RobRob
20219
$endgroup$
$begingroup$
Have you made any progress in getting to the desired equation? If so, would you be able to show us your working so far?
$endgroup$
– Minus One-Twelfth
Mar 23 at 6:12
add a comment |
$begingroup$
I am going over a physics text and I have difficulty to see how one can go from
$$2A = (1+ fracalphaik)(1+fracikalpha)fracFe^ikae^-alpha a2 + (1- fracalphaik)(1-fracikalpha)fracFe^ikae^alpha a2$$
to
$$fracAe^-ikaF = cosh(alpha a) + i(fracalpha^2 - k^22k alpha)sinh(alpha a)$$
Could you give me some hints on how to proceed? I, obviously, know $cosh(x) = frace^x+e^-x2$ and $sinh(x) = frace^x-e^-x2$ but the connection between the two steps is not obvious to me.
physics fractions
asked Mar 23 at 5:58
RobRob
20219
$endgroup$
I am going over a physics text and I have difficulty to see how one can go from
$$2A = (1+ fracalphaik)(1+fracikalpha)fracFe^ikae^-alpha a2 + (1- fracalphaik)(1-fracikalpha)fracFe^ikae^alpha a2$$
to
$$fracAe^-ikaF = cosh(alpha a) + i(fracalpha^2 - k^22k alpha)sinh(alpha a)$$
Could you give me some hints on how to proceed? I, obviously, know $cosh(x) = frace^x+e^-x2$ and $sinh(x) = frace^x-e^-x2$ but the connection between the two steps is not obvious to me.
physics fractions
physics fractions
asked Mar 23 at 5:58
RobRob
20219
asked Mar 23 at 5:58
RobRob
20219
asked Mar 23 at 5:58
RobRob
20219
asked Mar 23 at 5:58
RobRob
20219
asked Mar 23 at 5:58
RobRob
20219
20219
$begingroup$
Have you made any progress in getting to the desired equation? If so, would you be able to show us your working so far?
$endgroup$
– Minus One-Twelfth
Mar 23 at 6:12
add a comment |
$begingroup$
Have you made any progress in getting to the desired equation? If so, would you be able to show us your working so far?
$endgroup$
– Minus One-Twelfth
Mar 23 at 6:12
$begingroup$
Have you made any progress in getting to the desired equation? If so, would you be able to show us your working so far?
$endgroup$
– Minus One-Twelfth
Mar 23 at 6:12
$begingroup$
Have you made any progress in getting to the desired equation? If so, would you be able to show us your working so far?
$endgroup$
– Minus One-Twelfth
Mar 23 at 6:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Dividing by $2e^ikalphaF$ gives
$$fracAe^-ikalphaF=left(1+fracalphaikright)
left(1+fracikalpharight)frace^-aalpha4
+left(1-fracalphaikright)
left(1-fracikalpharight)frace^aalpha4.$$
Next,
$$left(1+fracalphaikright)
left(1+fracikalpharight)=frac(ik+alpha)^2ikalpha
=frac2kalpha+i(k^2-alpha^2)kalpha$$
and similarly
$$left(1-fracalphaikright)
left(1-fracikalpharight)=frac2kalpha-i(k^2-alpha^2)kalpha.$$
Then,
$$fracAe^-ikalphaF=frace^-aalpha+e^aalpha2
+ifrack^2-alpha^22kalphafrace^-aalpha-e^aalpha2
$$
which equals
$$cosh aalpha+ifracalpha^2-k^22kalphasinh aalpha.$$
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
add a comment |
$begingroup$
Hint: Multiply your first equation by $$frace^-ika2F$$
and expanding your paranthesis we get
$$left(2+fracalphaik+fracikalpharight)e^-alpha a+left(2-frac alphaik-fracikalpharight)e^alpha a$$
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Dividing by $2e^ikalphaF$ gives
$$fracAe^-ikalphaF=left(1+fracalphaikright)
left(1+fracikalpharight)frace^-aalpha4
+left(1-fracalphaikright)
left(1-fracikalpharight)frace^aalpha4.$$
Next,
$$left(1+fracalphaikright)
left(1+fracikalpharight)=frac(ik+alpha)^2ikalpha
=frac2kalpha+i(k^2-alpha^2)kalpha$$
and similarly
$$left(1-fracalphaikright)
left(1-fracikalpharight)=frac2kalpha-i(k^2-alpha^2)kalpha.$$
Then,
$$fracAe^-ikalphaF=frace^-aalpha+e^aalpha2
+ifrack^2-alpha^22kalphafrace^-aalpha-e^aalpha2
$$
which equals
$$cosh aalpha+ifracalpha^2-k^22kalphasinh aalpha.$$
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
add a comment |
$begingroup$
Dividing by $2e^ikalphaF$ gives
$$fracAe^-ikalphaF=left(1+fracalphaikright)
left(1+fracikalpharight)frace^-aalpha4
+left(1-fracalphaikright)
left(1-fracikalpharight)frace^aalpha4.$$
Next,
$$left(1+fracalphaikright)
left(1+fracikalpharight)=frac(ik+alpha)^2ikalpha
=frac2kalpha+i(k^2-alpha^2)kalpha$$
and similarly
$$left(1-fracalphaikright)
left(1-fracikalpharight)=frac2kalpha-i(k^2-alpha^2)kalpha.$$
Then,
$$fracAe^-ikalphaF=frace^-aalpha+e^aalpha2
+ifrack^2-alpha^22kalphafrace^-aalpha-e^aalpha2
$$
which equals
$$cosh aalpha+ifracalpha^2-k^22kalphasinh aalpha.$$
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
add a comment |
$begingroup$
Dividing by $2e^ikalphaF$ gives
$$fracAe^-ikalphaF=left(1+fracalphaikright)
left(1+fracikalpharight)frace^-aalpha4
+left(1-fracalphaikright)
left(1-fracikalpharight)frace^aalpha4.$$
Next,
$$left(1+fracalphaikright)
left(1+fracikalpharight)=frac(ik+alpha)^2ikalpha
=frac2kalpha+i(k^2-alpha^2)kalpha$$
and similarly
$$left(1-fracalphaikright)
left(1-fracikalpharight)=frac2kalpha-i(k^2-alpha^2)kalpha.$$
Then,
$$fracAe^-ikalphaF=frace^-aalpha+e^aalpha2
+ifrack^2-alpha^22kalphafrace^-aalpha-e^aalpha2
$$
which equals
$$cosh aalpha+ifracalpha^2-k^22kalphasinh aalpha.$$
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
Dividing by $2e^ikalphaF$ gives
$$fracAe^-ikalphaF=left(1+fracalphaikright)
left(1+fracikalpharight)frace^-aalpha4
+left(1-fracalphaikright)
left(1-fracikalpharight)frace^aalpha4.$$
Next,
$$left(1+fracalphaikright)
left(1+fracikalpharight)=frac(ik+alpha)^2ikalpha
=frac2kalpha+i(k^2-alpha^2)kalpha$$
and similarly
$$left(1-fracalphaikright)
left(1-fracikalpharight)=frac2kalpha-i(k^2-alpha^2)kalpha.$$
Then,
$$fracAe^-ikalphaF=frace^-aalpha+e^aalpha2
+ifrack^2-alpha^22kalphafrace^-aalpha-e^aalpha2
$$
which equals
$$cosh aalpha+ifracalpha^2-k^22kalphasinh aalpha.$$
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
edited Mar 23 at 6:32
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 23 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
add a comment |
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
$begingroup$
I think you forgot a factor of 2 in the denominator of the very last fraction. Other than that, it is perfect. Thanks!
$endgroup$
– Rob
Mar 23 at 6:31
add a comment |
$begingroup$
Hint: Multiply your first equation by $$frace^-ika2F$$
and expanding your paranthesis we get
$$left(2+fracalphaik+fracikalpharight)e^-alpha a+left(2-frac alphaik-fracikalpharight)e^alpha a$$
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: Multiply your first equation by $$frace^-ika2F$$
and expanding your paranthesis we get
$$left(2+fracalphaik+fracikalpharight)e^-alpha a+left(2-frac alphaik-fracikalpharight)e^alpha a$$
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: Multiply your first equation by $$frace^-ika2F$$
and expanding your paranthesis we get
$$left(2+fracalphaik+fracikalpharight)e^-alpha a+left(2-frac alphaik-fracikalpharight)e^alpha a$$
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
Hint: Multiply your first equation by $$frace^-ika2F$$
and expanding your paranthesis we get
$$left(2+fracalphaik+fracikalpharight)e^-alpha a+left(2-frac alphaik-fracikalpharight)e^alpha a$$
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
edited Mar 23 at 6:14
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered Mar 23 at 6:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
78.8k42867
add a comment |
add a comment |
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$begingroup$
Have you made any progress in getting to the desired equation? If so, would you be able to show us your working so far?
$endgroup$
– Minus One-Twelfth
Mar 23 at 6:12