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Recurrence relation of Bessel's and Legendre's function

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Partho Ghosh

This presentation tells about use recurrence relation in finding the solution of ordinary differential equations, with special emphasis on Bessel's and Legendre's Function.

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RECURRENCE RELATIONS (MATHEMATICS T.A. EC-IV SEMESTER) SUBJECT TEACHER: PROF. A.A. BASOLE GROUP MEMBERS 1. YASHWANT HAMPIHOLI (74) 2. YUVRAJ GUPTA (75) 3. PARTHO GHOSH (76) 4. ADARSH THAKUR (77) 5. AKSHAY PURWAR (78) 6. LUCKY THAKUR (79) 7. SHUBHAM SRIVASTAVA (80)
WHAT IS RECURRENCE RELATION? The concept of recurrence relations deals with recursive definitions of mathematical functions or sequences. Solving a recurrence relation involves, in finding "closed formβ€œ solution of the function. Recurrence relations are a fundamental mathematical tool since they can be used to represent mathematical functions/sequences that cannot be easily represented non-recursively. An example, is the Fibonacci sequence. Recurrence relations are largely employed in the design and analysis of algorithms.
RECURRENCE FORMULAE FOR BESSEL’S FUNCTION 𝑱 𝒏 𝒙 1. 𝒅 𝒙 𝒏 𝑱 𝒏(𝒙) 𝒅𝒙 = 𝒙 𝒏 𝑱 π’βˆ’πŸ(𝒙) 2. 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏(𝒙) 3. 𝑱 𝒏 𝒙 = 𝒙 πŸπ’ 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏(𝒙) 4. 𝑱 𝒏 β€² 𝒙 = 𝟏 𝟐 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) 5. 𝑱 𝒏 β€² 𝒙 = 𝒏 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) 6. 𝑱 𝒏+𝟏 𝒙 = πŸπ’ 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 π’βˆ’πŸ(𝒙)
PROOF FOR FIRST RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝒅 𝒙 𝒏 𝑱 𝒏(𝒙) 𝒅𝒙 = 𝒙 𝒏 𝑱 π’βˆ’πŸ(𝒙) Proof: Since, 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 𝒏+πŸπ’“ 𝟏 𝒓! πšͺ 𝐧 + 𝐫 + 𝟏 βˆ’ (𝟏) Multiplying equation (1) by π‘₯ 𝑛, we have 𝒙 𝒏 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 𝒏+𝒓 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 βˆ’ (𝟐)
CONTINUED… Differentiating equation (2) with respect to β€˜x’ on both sides ∴ 𝒅 𝒙 𝒏 𝑱 𝒏 𝒙 𝒅𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝟐 𝒏 + 𝒓 𝒙 𝟐 𝒏+𝒓 βˆ’πŸ 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 = 𝒙 𝒏 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 π’βˆ’πŸ+πŸπ’“ 𝒓! πšͺ 𝒏 βˆ’ 𝟏 + 𝒓 + 𝟏 = 𝒙 𝒏 𝑱 π’βˆ’πŸ 𝒙
PROOF FOR SECOND RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏(𝒙) Proof: Since, 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 𝒏+πŸπ’“ 𝟏 𝒓! πšͺ 𝐧 + 𝐫 + 𝟏 βˆ’ (𝟏) Multiplying equation (1) by π‘₯βˆ’π‘›, we have π’™βˆ’π’ 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 πŸπ’“ 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 βˆ’ 𝟐
CONTINUED… Differentiating equation (2) with respect to β€˜x’ on both sides ∴ 𝒅 π’™βˆ’π’ 𝑱 𝒏 𝒙 𝒅𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 πŸπ’“ 𝒙 πŸπ’“βˆ’πŸ 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 = βˆ’π’™βˆ’π’ 𝒓=𝟏 ∞ βˆ’πŸ π’“βˆ’πŸ 𝒙 𝒏+𝟏+𝟐 π’“βˆ’πŸ 𝟐 𝒏+𝟏+𝟐 π’“βˆ’πŸ 𝒓 βˆ’ 𝟏 ! πšͺ 𝒏 + 𝒓 + 𝟏 = βˆ’π’™βˆ’π’ π’Œ=𝟎 ∞ βˆ’πŸ π’Œ 𝒙 𝟐 𝒏+𝟏+πŸπ’Œ π’Œ! πšͺ 𝒏 + 𝟏 + π’Œ + 𝟏 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏 𝒙 , π’˜π’‰π’†π’“π’† π’Œ = 𝒓 βˆ’ 𝟏
PROOF FOR THIRD RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝒙 πŸπ’ 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏(𝒙) Proof: Since, 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏 𝒙 … (πŸ’) On differentiating both sides of equation (1) with respect to β€˜x’: 𝒙 𝒏 𝑱 𝒏 β€² 𝒙 + 𝒏𝒙 π’βˆ’πŸ 𝑱 𝒏 𝒙 = 𝒙 𝒏 𝑱 π’βˆ’πŸ 𝒙 … (𝟐)
CONTINUED… On dividing equation (2) by π‘₯ 𝑛 : 𝑱 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 … (πŸ‘) Since, 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏 𝒙 … (πŸ’) On differentiating both sides of equation (4) with respect to β€˜x’: βˆ’π‘± 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 𝒏+𝟏 𝒙 … πŸ“
CONTINUED… πŸπ’ 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏 𝒙 … (πŸ”) i.e., 𝑱 𝒏 𝒙 = 𝒙 πŸπ’ 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏(𝒙) … (πŸ•)
PROOF FOR FOURTH RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏 β€² 𝒙 = 𝟏 𝟐 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) Proof: Since we know that βˆ’π‘± 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 𝒏+𝟏 𝒙 … 𝟏 And 𝑱 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 … (𝟐)
CONTINUED… So on subtracting equation (1) from (2), we get πŸπ‘± 𝒏 β€² 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏 𝒙 … πŸ‘ i.e. , 𝑱 𝒏 β€² 𝒙 = 𝟏 𝟐 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) … (πŸ—)
PROOF FOR FIFTH RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏 β€² 𝒙 = 𝒏 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) Proof: Since we know that βˆ’π‘± 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 𝒏+𝟏 𝒙 … 𝟏 The equation (1) can also be represented as: 𝑱 𝒏 β€² 𝒙 = 𝒏 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 𝒏+𝟏 𝒙
PROOF FOR SIXTH RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏+𝟏 𝒙 = πŸπ’ 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 π’βˆ’πŸ(𝒙) Proof: Since we know that πŸπ’ 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏 𝒙 … (𝟏) The equation (1) can also be represented as: πŸπ’ 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏 𝒙
RECURRENCE FORMULAE FOR 𝑷 𝒏 𝒙 1. 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ(𝒙) 2. 𝒏𝑷 𝒏 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) 3. πŸπ’ + 𝟏 𝑷 𝒏 𝒙 = 𝑷 𝒏+𝟏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) 4. 𝑷 𝒏 β€² 𝒙 = 𝒙𝑷 π’βˆ’πŸ β€² 𝒙 + 𝒏𝑷 π’βˆ’πŸ(𝒙) 5. 𝟏 βˆ’ 𝒙 𝟐 𝑷 𝒏 β€² 𝒙 = 𝒏[𝑷 π’βˆ’πŸ 𝒙 βˆ’ 𝒙𝑷 𝒏(𝒙)]
PROOF FOR FIRST RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ(𝒙) Proof: We know that (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 )βˆ’ 𝟏 𝟐 = 𝒏=𝟎 ∞ 𝑷 𝒏 𝒙 𝒕 𝒏 … (𝟏) Differentiating (1) partially w.r.t. t, we get βˆ’ 𝟏 𝟐 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 βˆ’πŸπ’™ + πŸπ’• = 𝒏𝑷 𝒏 (𝒙)𝒕 π’βˆ’πŸ Or 𝒙 βˆ’ 𝒕 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 = (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 ) 𝒏𝑷 𝒏(𝒙)𝒕 π’βˆ’πŸ Or 𝒙 βˆ’ 𝒕 𝑷 𝒏 𝒙 𝒕 𝒏 = (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 ) 𝒏𝑷 𝒏 𝒙 𝒕 π’βˆ’πŸ Equating coefficients of 𝑑 𝑛 from both sides, we get 𝒙𝑷 𝒏 𝒙 βˆ’ 𝑷 π’βˆ’πŸ 𝒙 = 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 βˆ’ πŸπ’π’™π‘· 𝒏 𝒙 + (𝒏 βˆ’ 𝟏)𝑷 π’βˆ’πŸ(𝒙)
PROOF FOR SECOND RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝒏𝑷 𝒏 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) Proof: We know that (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 )βˆ’ 𝟏 𝟐 = 𝒏=𝟎 ∞ 𝑷 𝒏 𝒙 𝒕 𝒏 … (𝟏) Differentiating (1) partially w.r.t. x, we get βˆ’ 𝟏 𝟐 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 . πŸπ’• = 𝑷 𝒏′(𝒙)𝒕 𝒏 i.e., 𝒕(𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 )βˆ’ πŸ‘ 𝟐 = 𝑷 𝒏 β€² 𝒙 𝒕 𝒏 …(2) Again differentiating (1) partially w.r.t. t, we have 𝒙 βˆ’ 𝒕 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 = 𝒏𝑷 𝒏 𝒙 𝒕 π’βˆ’πŸ …(3) Dividing (3) by (2), we get 𝒙 βˆ’ 𝒕 𝒕 = 𝒏 𝑷 𝒏(𝒙)𝒕 π’βˆ’πŸ 𝑷 𝒏 β€²(𝒙)𝒕 𝒏 i.e. 𝒏𝑷 𝒏 𝒙 𝒕 𝒏 = (𝒙 βˆ’ 𝒕) 𝑷 𝒏′ 𝒙 𝒕 𝒏 …(4) Equating coefficient of 𝑑 𝑛 from both sides of equation (4) we get: 𝒏𝑷 𝒏 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙)
PROOF FOR THIRD RECURRENCE RELATION OF 𝑷 𝒏(𝒙) πŸπ’ + 𝟏 𝑷 𝒏 𝒙 = 𝑷 𝒏+𝟏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) Proof: Since we know that: 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ 𝒙 … (𝟏) Differentiating (1) w.r.t. x, we get 𝒏 + 𝟏 𝑷 𝒏+𝟏 β€² 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + πŸπ’ + 𝟏 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ β€² 𝒙 … 𝟐 Substituting for nπ‘₯𝑃𝑛 β€² π‘₯ π‘“π‘Ÿπ‘œπ‘š π‘ π‘’π‘π‘œπ‘›π‘‘ π‘Ÿπ‘’π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘π‘’ π‘Ÿπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝑖𝑛 (2), we obtain 𝒏 + 𝟏 𝑷 𝒏+𝟏 β€² 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 + 𝑷 π’βˆ’πŸ β€² 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸβ€²(𝒙) Or 𝟐𝐧 + 𝟏 𝑷 𝒏 𝒙 = 𝑷 𝒏+πŸβ€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸβ€²(𝒙)
PROOF FOR FOURTH RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝑷 𝒏 β€² 𝒙 = 𝒙𝑷 π’βˆ’πŸ β€² 𝒙 + 𝒏𝑷 π’βˆ’πŸ 𝒙 Proof: Since we know that: 𝒏 + 𝟏 𝑷 𝒏+𝟏 β€² 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + πŸπ’ + 𝟏 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ β€² 𝒙 … (𝟏) Rewriting (1) as: 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + 𝒏 + 𝟏 𝒙𝑷 𝒏 , 𝒙 + 𝒏 𝒙𝑷 𝒏 , 𝒙 βˆ’ 𝑷 π’βˆ’πŸ , 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + 𝒏 + 𝟏 𝒙𝑷 𝒏 , 𝒙 + 𝒏 𝟐 𝑷 𝒏(𝒙) = 𝒏 + 𝟏 𝒙𝑷 𝒏 , 𝒙 + 𝒏 𝟐 + πŸπ’ + 𝟏 𝑷 𝒏 𝒙 Or 𝑷 𝒏+𝟏 β€² 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 + (𝒏 + 𝟏)𝑷 𝒙 …(2) Replacing n by (n-1) in equation (2), we get the required equation
PROOF FOR FIFTH RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝟏 βˆ’ 𝒙 𝟐 𝑷 𝒏 β€² 𝒙 = 𝒏[𝑷 π’βˆ’πŸ 𝒙 βˆ’ 𝒙𝑷 𝒏(𝒙)] Proof: Rewriting second and fourth recurrence relation of 𝑃𝑛 π‘₯ as: 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² 𝒙 = 𝒏𝑷 𝒏 𝒙 …(1) and 𝑷 𝒏 β€² 𝒙 βˆ’ 𝒙𝑷 π’βˆ’πŸ β€² 𝒙 = 𝒏𝑷 π’βˆ’πŸ 𝒙 …(2) Multiplying (1) by x and subtracting from (2), we get 𝟏 βˆ’ 𝒙 𝟐 𝑷 𝒏 β€² 𝒙 = 𝒏[𝑷 π’βˆ’πŸ 𝒙 βˆ’ 𝒙𝑷 𝒏(𝒙)]
Recurrence relation of Bessel's and Legendre's function

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Recurrence relation of Bessel's and Legendre's function

  • 1. RECURRENCE RELATIONS (MATHEMATICS T.A. EC-IV SEMESTER) SUBJECT TEACHER: PROF. A.A. BASOLE GROUP MEMBERS 1. YASHWANT HAMPIHOLI (74) 2. YUVRAJ GUPTA (75) 3. PARTHO GHOSH (76) 4. ADARSH THAKUR (77) 5. AKSHAY PURWAR (78) 6. LUCKY THAKUR (79) 7. SHUBHAM SRIVASTAVA (80)
  • 2. WHAT IS RECURRENCE RELATION? The concept of recurrence relations deals with recursive definitions of mathematical functions or sequences. Solving a recurrence relation involves, in finding "closed formβ€œ solution of the function. Recurrence relations are a fundamental mathematical tool since they can be used to represent mathematical functions/sequences that cannot be easily represented non-recursively. An example, is the Fibonacci sequence. Recurrence relations are largely employed in the design and analysis of algorithms.
  • 3. RECURRENCE FORMULAE FOR BESSEL’S FUNCTION 𝑱 𝒏 𝒙 1. 𝒅 𝒙 𝒏 𝑱 𝒏(𝒙) 𝒅𝒙 = 𝒙 𝒏 𝑱 π’βˆ’πŸ(𝒙) 2. 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏(𝒙) 3. 𝑱 𝒏 𝒙 = 𝒙 πŸπ’ 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏(𝒙) 4. 𝑱 𝒏 β€² 𝒙 = 𝟏 𝟐 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) 5. 𝑱 𝒏 β€² 𝒙 = 𝒏 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) 6. 𝑱 𝒏+𝟏 𝒙 = πŸπ’ 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 π’βˆ’πŸ(𝒙)
  • 4. PROOF FOR FIRST RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝒅 𝒙 𝒏 𝑱 𝒏(𝒙) 𝒅𝒙 = 𝒙 𝒏 𝑱 π’βˆ’πŸ(𝒙) Proof: Since, 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 𝒏+πŸπ’“ 𝟏 𝒓! πšͺ 𝐧 + 𝐫 + 𝟏 βˆ’ (𝟏) Multiplying equation (1) by π‘₯ 𝑛, we have 𝒙 𝒏 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 𝒏+𝒓 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 βˆ’ (𝟐)
  • 5. CONTINUED… Differentiating equation (2) with respect to β€˜x’ on both sides ∴ 𝒅 𝒙 𝒏 𝑱 𝒏 𝒙 𝒅𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝟐 𝒏 + 𝒓 𝒙 𝟐 𝒏+𝒓 βˆ’πŸ 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 = 𝒙 𝒏 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 π’βˆ’πŸ+πŸπ’“ 𝒓! πšͺ 𝒏 βˆ’ 𝟏 + 𝒓 + 𝟏 = 𝒙 𝒏 𝑱 π’βˆ’πŸ 𝒙
  • 6. PROOF FOR SECOND RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏(𝒙) Proof: Since, 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 𝟐 𝒏+πŸπ’“ 𝟏 𝒓! πšͺ 𝐧 + 𝐫 + 𝟏 βˆ’ (𝟏) Multiplying equation (1) by π‘₯βˆ’π‘›, we have π’™βˆ’π’ 𝑱 𝒏 𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 𝒙 πŸπ’“ 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 βˆ’ 𝟐
  • 7. CONTINUED… Differentiating equation (2) with respect to β€˜x’ on both sides ∴ 𝒅 π’™βˆ’π’ 𝑱 𝒏 𝒙 𝒅𝒙 = 𝒓=𝟎 ∞ βˆ’πŸ 𝒓 πŸπ’“ 𝒙 πŸπ’“βˆ’πŸ 𝟐 𝒏+πŸπ’“ 𝒓! πšͺ 𝒏 + 𝒓 + 𝟏 = βˆ’π’™βˆ’π’ 𝒓=𝟏 ∞ βˆ’πŸ π’“βˆ’πŸ 𝒙 𝒏+𝟏+𝟐 π’“βˆ’πŸ 𝟐 𝒏+𝟏+𝟐 π’“βˆ’πŸ 𝒓 βˆ’ 𝟏 ! πšͺ 𝒏 + 𝒓 + 𝟏 = βˆ’π’™βˆ’π’ π’Œ=𝟎 ∞ βˆ’πŸ π’Œ 𝒙 𝟐 𝒏+𝟏+πŸπ’Œ π’Œ! πšͺ 𝒏 + 𝟏 + π’Œ + 𝟏 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏 𝒙 , π’˜π’‰π’†π’“π’† π’Œ = 𝒓 βˆ’ 𝟏
  • 8. PROOF FOR THIRD RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝒙 πŸπ’ 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏(𝒙) Proof: Since, 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏 𝒙 … (πŸ’) On differentiating both sides of equation (1) with respect to β€˜x’: 𝒙 𝒏 𝑱 𝒏 β€² 𝒙 + 𝒏𝒙 π’βˆ’πŸ 𝑱 𝒏 𝒙 = 𝒙 𝒏 𝑱 π’βˆ’πŸ 𝒙 … (𝟐)
  • 9. CONTINUED… On dividing equation (2) by π‘₯ 𝑛 : 𝑱 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 … (πŸ‘) Since, 𝒅 π’™βˆ’π’ 𝑱 𝒏(𝒙) 𝒅𝒙 = βˆ’π’™βˆ’π’ 𝑱 𝒏+𝟏 𝒙 … (πŸ’) On differentiating both sides of equation (4) with respect to β€˜x’: βˆ’π‘± 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 𝒏+𝟏 𝒙 … πŸ“
  • 10. CONTINUED… πŸπ’ 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏 𝒙 … (πŸ”) i.e., 𝑱 𝒏 𝒙 = 𝒙 πŸπ’ 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏(𝒙) … (πŸ•)
  • 11. PROOF FOR FOURTH RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏 β€² 𝒙 = 𝟏 𝟐 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) Proof: Since we know that βˆ’π‘± 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 𝒏+𝟏 𝒙 … 𝟏 And 𝑱 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 … (𝟐)
  • 12. CONTINUED… So on subtracting equation (1) from (2), we get πŸπ‘± 𝒏 β€² 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏 𝒙 … πŸ‘ i.e. , 𝑱 𝒏 β€² 𝒙 = 𝟏 𝟐 𝑱 π’βˆ’πŸ 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) … (πŸ—)
  • 13. PROOF FOR FIFTH RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏 β€² 𝒙 = 𝒏 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 𝒏+𝟏(𝒙) Proof: Since we know that βˆ’π‘± 𝒏 β€² 𝒙 + 𝒏 𝒙 𝑱 𝒏 𝒙 = 𝑱 𝒏+𝟏 𝒙 … 𝟏 The equation (1) can also be represented as: 𝑱 𝒏 β€² 𝒙 = 𝒏 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 𝒏+𝟏 𝒙
  • 14. PROOF FOR SIXTH RECURRENCE RELATION OF 𝑱 𝒏 𝒙 𝑱 𝒏+𝟏 𝒙 = πŸπ’ 𝒙 𝑱 𝒏 𝒙 βˆ’ 𝑱 π’βˆ’πŸ(𝒙) Proof: Since we know that πŸπ’ 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏 𝒙 … (𝟏) The equation (1) can also be represented as: πŸπ’ 𝒙 𝑱 𝒏 𝒙 = 𝑱 π’βˆ’πŸ 𝒙 + 𝑱 𝒏+𝟏 𝒙
  • 15. RECURRENCE FORMULAE FOR 𝑷 𝒏 𝒙 1. 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ(𝒙) 2. 𝒏𝑷 𝒏 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) 3. πŸπ’ + 𝟏 𝑷 𝒏 𝒙 = 𝑷 𝒏+𝟏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) 4. 𝑷 𝒏 β€² 𝒙 = 𝒙𝑷 π’βˆ’πŸ β€² 𝒙 + 𝒏𝑷 π’βˆ’πŸ(𝒙) 5. 𝟏 βˆ’ 𝒙 𝟐 𝑷 𝒏 β€² 𝒙 = 𝒏[𝑷 π’βˆ’πŸ 𝒙 βˆ’ 𝒙𝑷 𝒏(𝒙)]
  • 16. PROOF FOR FIRST RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ(𝒙) Proof: We know that (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 )βˆ’ 𝟏 𝟐 = 𝒏=𝟎 ∞ 𝑷 𝒏 𝒙 𝒕 𝒏 … (𝟏) Differentiating (1) partially w.r.t. t, we get βˆ’ 𝟏 𝟐 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 βˆ’πŸπ’™ + πŸπ’• = 𝒏𝑷 𝒏 (𝒙)𝒕 π’βˆ’πŸ Or 𝒙 βˆ’ 𝒕 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 = (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 ) 𝒏𝑷 𝒏(𝒙)𝒕 π’βˆ’πŸ Or 𝒙 βˆ’ 𝒕 𝑷 𝒏 𝒙 𝒕 𝒏 = (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 ) 𝒏𝑷 𝒏 𝒙 𝒕 π’βˆ’πŸ Equating coefficients of 𝑑 𝑛 from both sides, we get 𝒙𝑷 𝒏 𝒙 βˆ’ 𝑷 π’βˆ’πŸ 𝒙 = 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 βˆ’ πŸπ’π’™π‘· 𝒏 𝒙 + (𝒏 βˆ’ 𝟏)𝑷 π’βˆ’πŸ(𝒙)
  • 17. PROOF FOR SECOND RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝒏𝑷 𝒏 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) Proof: We know that (𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 )βˆ’ 𝟏 𝟐 = 𝒏=𝟎 ∞ 𝑷 𝒏 𝒙 𝒕 𝒏 … (𝟏) Differentiating (1) partially w.r.t. x, we get βˆ’ 𝟏 𝟐 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 . πŸπ’• = 𝑷 𝒏′(𝒙)𝒕 𝒏 i.e., 𝒕(𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 )βˆ’ πŸ‘ 𝟐 = 𝑷 𝒏 β€² 𝒙 𝒕 𝒏 …(2) Again differentiating (1) partially w.r.t. t, we have 𝒙 βˆ’ 𝒕 𝟏 βˆ’ πŸπ’™π’• + 𝒕 𝟐 βˆ’ πŸ‘ 𝟐 = 𝒏𝑷 𝒏 𝒙 𝒕 π’βˆ’πŸ …(3) Dividing (3) by (2), we get 𝒙 βˆ’ 𝒕 𝒕 = 𝒏 𝑷 𝒏(𝒙)𝒕 π’βˆ’πŸ 𝑷 𝒏 β€²(𝒙)𝒕 𝒏 i.e. 𝒏𝑷 𝒏 𝒙 𝒕 𝒏 = (𝒙 βˆ’ 𝒕) 𝑷 𝒏′ 𝒙 𝒕 𝒏 …(4) Equating coefficient of 𝑑 𝑛 from both sides of equation (4) we get: 𝒏𝑷 𝒏 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙)
  • 18. PROOF FOR THIRD RECURRENCE RELATION OF 𝑷 𝒏(𝒙) πŸπ’ + 𝟏 𝑷 𝒏 𝒙 = 𝑷 𝒏+𝟏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² (𝒙) Proof: Since we know that: 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ 𝒙 … (𝟏) Differentiating (1) w.r.t. x, we get 𝒏 + 𝟏 𝑷 𝒏+𝟏 β€² 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + πŸπ’ + 𝟏 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ β€² 𝒙 … 𝟐 Substituting for nπ‘₯𝑃𝑛 β€² π‘₯ π‘“π‘Ÿπ‘œπ‘š π‘ π‘’π‘π‘œπ‘›π‘‘ π‘Ÿπ‘’π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘π‘’ π‘Ÿπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝑖𝑛 (2), we obtain 𝒏 + 𝟏 𝑷 𝒏+𝟏 β€² 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + πŸπ’ + 𝟏 𝒙𝑷 𝒏 𝒙 + 𝑷 π’βˆ’πŸ β€² 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸβ€²(𝒙) Or 𝟐𝐧 + 𝟏 𝑷 𝒏 𝒙 = 𝑷 𝒏+πŸβ€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸβ€²(𝒙)
  • 19. PROOF FOR FOURTH RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝑷 𝒏 β€² 𝒙 = 𝒙𝑷 π’βˆ’πŸ β€² 𝒙 + 𝒏𝑷 π’βˆ’πŸ 𝒙 Proof: Since we know that: 𝒏 + 𝟏 𝑷 𝒏+𝟏 β€² 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + πŸπ’ + 𝟏 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝒏𝑷 π’βˆ’πŸ β€² 𝒙 … (𝟏) Rewriting (1) as: 𝒏 + 𝟏 𝑷 𝒏+𝟏 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + 𝒏 + 𝟏 𝒙𝑷 𝒏 , 𝒙 + 𝒏 𝒙𝑷 𝒏 , 𝒙 βˆ’ 𝑷 π’βˆ’πŸ , 𝒙 = πŸπ’ + 𝟏 𝑷 𝒏 𝒙 + 𝒏 + 𝟏 𝒙𝑷 𝒏 , 𝒙 + 𝒏 𝟐 𝑷 𝒏(𝒙) = 𝒏 + 𝟏 𝒙𝑷 𝒏 , 𝒙 + 𝒏 𝟐 + πŸπ’ + 𝟏 𝑷 𝒏 𝒙 Or 𝑷 𝒏+𝟏 β€² 𝒙 = 𝒙𝑷 𝒏 β€² 𝒙 + (𝒏 + 𝟏)𝑷 𝒙 …(2) Replacing n by (n-1) in equation (2), we get the required equation
  • 20. PROOF FOR FIFTH RECURRENCE RELATION OF 𝑷 𝒏(𝒙) 𝟏 βˆ’ 𝒙 𝟐 𝑷 𝒏 β€² 𝒙 = 𝒏[𝑷 π’βˆ’πŸ 𝒙 βˆ’ 𝒙𝑷 𝒏(𝒙)] Proof: Rewriting second and fourth recurrence relation of 𝑃𝑛 π‘₯ as: 𝒙𝑷 𝒏 β€² 𝒙 βˆ’ 𝑷 π’βˆ’πŸ β€² 𝒙 = 𝒏𝑷 𝒏 𝒙 …(1) and 𝑷 𝒏 β€² 𝒙 βˆ’ 𝒙𝑷 π’βˆ’πŸ β€² 𝒙 = 𝒏𝑷 π’βˆ’πŸ 𝒙 …(2) Multiplying (1) by x and subtracting from (2), we get 𝟏 βˆ’ 𝒙 𝟐 𝑷 𝒏 β€² 𝒙 = 𝒏[𝑷 π’βˆ’πŸ 𝒙 βˆ’ 𝒙𝑷 𝒏(𝒙)]