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orthogonal chebyshev polyhomials

Definite integral including the Chebyshev polynomialUsing the half/double angle formulas to solve an equationProve that $T_n$ satisfy $ sum_{k=0}^{N-1}{T_i(x_k)T_j(x_k)} = begin{cases} 0 &: ine j \ lneq 0 &: i=j end{cases} ,! $Building a cubic function with integer coefficients and trigonometric rootsHow can I get the value for $theta$ when $cos(theta) = frac{sqrt{3}-1}{2sqrt2}$, $sin(theta) = frac{sqrt3+1}{2sqrt2}$Triple integral - converting to cylindrical coordinatesCompute $int_{-1}^{1} T_n(x) T_m(x) frac{dx}{sqrt{1-x^2}}$On the extrema of Chebyshev polynomials of the second kindIntegrate orthogonal function over solid angleRoot of cosine function

1

$begingroup$

This Theorem says

statement:
$int_{-1}^{1} frac{T_n(x)T_m(x)}{ sqrt{1+x^2} } dx = 0 ;$ when $nne m $

proof:
"substitute $x= cos theta$ "
and that's it.

So I am wondering should I start with this

$int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+x^2}} dx $
or with this

$int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+cos^2 theta}} dx $
I order to verify the proof.

note: T(x) is Chebyshev polynomial.

calculus algebra-precalculus

share|cite|improve this question

edited Mar 10 at 3:50

tt z

asked Mar 10 at 3:32

tt ztt z

265

$endgroup$

add a comment | 

1

$begingroup$

This Theorem says

statement:
$int_{-1}^{1} frac{T_n(x)T_m(x)}{ sqrt{1+x^2} } dx = 0 ;$ when $nne m $

proof:
"substitute $x= cos theta$ "
and that's it.

So I am wondering should I start with this

$int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+x^2}} dx $
or with this

$int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+cos^2 theta}} dx $
I order to verify the proof.

note: T(x) is Chebyshev polynomial.

calculus algebra-precalculus

share|cite|improve this question

edited Mar 10 at 3:50

tt z

asked Mar 10 at 3:32

tt ztt z

265

$endgroup$

add a comment | 

$begingroup$

This Theorem says

statement:
$int_{-1}^{1} frac{T_n(x)T_m(x)}{ sqrt{1+x^2} } dx = 0 ;$ when $nne m $

proof:
"substitute $x= cos theta$ "
and that's it.

So I am wondering should I start with this

$int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+x^2}} dx $
or with this

$int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+cos^2 theta}} dx $
I order to verify the proof.

note: T(x) is Chebyshev polynomial.

calculus algebra-precalculus

share|cite|improve this question

edited Mar 10 at 3:50

tt z

asked Mar 10 at 3:32

tt ztt z

265

$endgroup$

share|cite|improve this question

edited Mar 10 at 3:50

tt z

asked Mar 10 at 3:32

tt ztt z

265

share|cite|improve this question

edited Mar 10 at 3:50

tt z

asked Mar 10 at 3:32

tt ztt z

265

asked Mar 10 at 3:32

tt ztt z

265

asked Mar 10 at 3:32

tt ztt z

265

1 Answer
1

active

oldest

votes

1

$begingroup$

When you do this, you apply the substitution rule for integrals, fully. Everything with $x$ in it, including the denominator, the $dx$, and the limits, transform.

On the other hand, the statement that you're trying to prove is incorrect. The correct form has $sqrt{1-x^2}$ in the denominator instead.

share|cite|improve this answer

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+cos^2 theta}} dx = int_{0}^{pi} cos(theta m). cos(theta n) dtheta$ I wonder how this come about when we change $x$ to $cos(theta)$ where does the denominator go.
  $endgroup$
  – tt z
  Mar 10 at 4:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It goes into the $dtheta$. Well, it does if you use the correct version with a minus sign in that square root.
  $endgroup$
  – jmerry
  Mar 10 at 4:44
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, trig substitutions.
  $endgroup$
  – tt z
  Mar 10 at 4:57
  

  
  
  
  

add a comment | 

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1

$begingroup$

When you do this, you apply the substitution rule for integrals, fully. Everything with $x$ in it, including the denominator, the $dx$, and the limits, transform.

On the other hand, the statement that you're trying to prove is incorrect. The correct form has $sqrt{1-x^2}$ in the denominator instead.

share|cite|improve this answer

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+cos^2 theta}} dx = int_{0}^{pi} cos(theta m). cos(theta n) dtheta$ I wonder how this come about when we change $x$ to $cos(theta)$ where does the denominator go.
  $endgroup$
  – tt z
  Mar 10 at 4:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It goes into the $dtheta$. Well, it does if you use the correct version with a minus sign in that square root.
  $endgroup$
  – jmerry
  Mar 10 at 4:44
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, trig substitutions.
  $endgroup$
  – tt z
  Mar 10 at 4:57
  

  
  
  
  

add a comment | 

1

$begingroup$

When you do this, you apply the substitution rule for integrals, fully. Everything with $x$ in it, including the denominator, the $dx$, and the limits, transform.

On the other hand, the statement that you're trying to prove is incorrect. The correct form has $sqrt{1-x^2}$ in the denominator instead.

share|cite|improve this answer

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $int_{-1}^{1} frac{cos(theta n)cos(theta m)}{sqrt{1+cos^2 theta}} dx = int_{0}^{pi} cos(theta m). cos(theta n) dtheta$ I wonder how this come about when we change $x$ to $cos(theta)$ where does the denominator go.
  $endgroup$
  – tt z
  Mar 10 at 4:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It goes into the $dtheta$. Well, it does if you use the correct version with a minus sign in that square root.
  $endgroup$
  – jmerry
  Mar 10 at 4:44
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, trig substitutions.
  $endgroup$
  – tt z
  Mar 10 at 4:57
  

  
  
  
  

add a comment | 

$begingroup$

When you do this, you apply the substitution rule for integrals, fully. Everything with $x$ in it, including the denominator, the $dx$, and the limits, transform.

On the other hand, the statement that you're trying to prove is incorrect. The correct form has $sqrt{1-x^2}$ in the denominator instead.

share|cite|improve this answer

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629

$endgroup$

share|cite|improve this answer

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629

answered Mar 10 at 3:42

jmerryjmerry

13.8k1629